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#### Question 1:

Prove the following trigonometric identities.

(1 − cos2 A) cosec2 A = 1

We know that,

So,

#### Question 2:

Prove the following trigonometric identities.

(1 + cot2 A) sin2 A = 1

We know that,

So,

#### Question 3:

Prove the following trigonometric identities.

tan2θ cos2θ = 1 − cos2θ

We know that, .

So,

#### Question 4:

Prove the following trigonometric identities.

We know that,

So,

#### Question 5:

Prove the following trigonometric identities.

(sec2 θ − 1) (cosec2 θ − 1) = 1

We know that,

So,

#### Question 6:

Prove the following trigonometric identities.

We know that,

So,

#### Question 7:

Prove the following trigonometric identities.

We know that,

Multiplying both numerator and the denominator by, we have

#### Question 8:

Prove the following trigonometric identities.

We know that,

Multiplying the both numerator and the denominator by, we have

#### Question 9:

Prove the following trigonometric identities.

We know that,

So,

#### Question 10:

Prove the following trigonometric identities.

We know that,

So,

#### Question 11:

Prove the following trigonometric identities.

We know that,

Multiplying numerator and denominator under the square root by , we have

#### Question 12:

Prove the following trigonometric identities.

We have to prove .

We know that,

Multiplying both numerator and denominator by , we have

#### Question 13:

Prove the following trigonometric identities.

We have to prove .

We know that,

Multiplying both numerator and denominator by , we have

#### Question 14:

Prove the following trigonometric identities.

We have to prove

We know that, .

Multiplying both numerator and denominator by , we have

#### Question 15:

Prove the following trigonometric identities.

We have to prove

We know that,

So,

#### Question 16:

Prove the following trigonometric identities.

tan2θ − sin2θ = tan2θ sin2θ

We have to prove

We know that,

So,

#### Question 17:

Prove the following trigonometric identities.

(cosecθ + sinθ) (cosecθ − sinθ) = cot2 θ + cos2θ

We have to prove

We know that,

So,

#### Question 18:

Prove the following trigonometric identities.

(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ

We have to prove

We know that,

#### Question 19:

Prove the following trigonometric identities.

secA (1 − sinA) (secA + tanA) = 1

We have to prove

We know that,

So,

#### Question 20:

Prove the following trigonometric identities.

(cosecA − sinA) (secA − cosA) (tanA + cotA) = 1

We have to prove

We know that,

So,

#### Question 21:

Prove the following trigonometric identities.

(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1

We have to prove

We know that,

So,

#### Question 22:

Prove the following trigonometric identities.

sin2 A cot2 A + cos2 A tan2 A = 1

We have to prove

We know that,

So,

#### Question 23:

Prove the following trigonometric identities.

(i)

(ii)

(iii)

(i) We have to prove

We know that,

So,

(ii) We have to prove

We know that,

So,

(iii)
Given

Hence proved.

#### Question 24:

Prove the following trigonometric identities.

We have to prove

We know that,

So,

#### Question 25:

Prove the following trigonometric identities.

We have to prove

We know that,

So,

#### Question 26:

Prove the following trigonometric identities.

We have to prove

We know that,

Multiplying the denominator and numerator of the second term by , we have

#### Question 27:

Prove the following trigonometric identities.

We have to prove that .

We know that,

So,

#### Question 28:

Prove the following trigonometric identities.

$\frac{1+{\mathrm{tan}}^{2}\theta }{1+{\mathrm{cot}}^{2}\theta }={\left(\frac{1-\mathrm{tan}\theta }{1-\mathrm{cot}\theta }\right)}^{2}={\mathrm{tan}}^{2}\theta$

We have to prove

Consider the expression

Again, we have

#### Question 29:

Prove the following trigonometric identities.

$\frac{1+\mathrm{sec}\theta }{\mathrm{sec}\theta }=\frac{{\mathrm{sin}}^{2}\theta }{1-\mathrm{cos}\theta }$

We have to prove

We know that,

Multiplying the numerator and denominator by , we have

#### Question 30:

Prove the following trigonometric identities.

$\frac{\mathrm{tan}\theta }{1-\mathrm{cot}\theta }+\frac{\mathrm{cot}\theta }{1-\mathrm{tan}\theta }=1+\mathrm{tan}\theta +\mathrm{cot}\theta$

We need to prove

Now, using in the L.H.S, we get

Further using the identity, we get

Hence

#### Question 31:

Prove the following trigonometric identities.

sec6θ = tan6θ + 3 tan2θ sec2θ + 1

We need to prove

Solving the L.H.S, we get

Further using the identity, we get

Hence proved.

#### Question 32:

Prove the following trigonometric identities

cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1

We need to prove

Solving the L.H.S, we get

()

Further using the identity, we get

Hence proved.

#### Question 33:

Prove the following trigonometric identities.

We need to prove

Solving the L.H.S, we get

Using , we get

Hence proved.

#### Question 34:

Prove the following trigonometric identities.

$\frac{1+\mathrm{cos}A}{{\mathrm{sin}}^{2}A}=\frac{1}{1-\mathrm{cos}A}$

We need to prove

Using the property , we get

LHS =

Further using the identity, , we get

= RHS

Hence proved.

#### Question 35:

Prove the following trigonometric identities.

$\frac{\mathrm{sec}A-\mathrm{tan}A}{\mathrm{sec}A+\mathrm{tan}A}=\frac{{\mathrm{cos}}^{2}A}{{\left(1+\mathrm{sin}A\right)}^{2}}$

We need to prove

Here, we will first solve the LHS.

Now, using , we get

Further, multiplying both numerator and denominator by , we get

Now, using the property, we get

So,

= RHS

Hence proved.

#### Question 36:

Prove the following trigonometric identities.

We need to prove

Now, multiplying the numerator and denominator of LHS by , we get

Further using the identity, , we get

Hence proved.

#### Question 37:

Prove the following trigonometric identities.

(i)

(ii)

(i) We need to prove

Here, rationalising the L.H.S, we get

Further using the property, , we get

So,

Hence proved.

(ii) We need to prove

Here, rationaliaing the L.H.S, we get

Further using the property, , we get

So,

Hence proved.

#### Question 38:

Prove the following trigonometric identities.

(i)
(ii)
(iii)
(iv)
(v) $\frac{\mathrm{sin}\theta +1-\mathrm{cos}\theta }{\mathrm{cos}\theta -1+\mathrm{sin}\theta }=\frac{1+\mathrm{sin}\theta }{\mathrm{cos}\theta }$

(i) We have,

(ii) We have,

(iii) We have,

(iv) We have,

Multiplying both the numerator and the denominator by, we have

(v) We have,

Multiplying both the numerator and the denominator by , we have

#### Question 39:

Prove the following trigonometric identities.

We need to prove

Here, we will first solve the L.H.S.

Now, using , we get

Further using the property , we get

So,

Hence proved.

#### Question 40:

Prove the following trigonometric identities.

We need to prove

Now, rationalising the L.H.S, we get

Using and, we get

Hence proved.

#### Question 41:

Prove the following trigonometric identities.

We need to prove

Solving the L.H.S, we get

Further using the property , we get

So,

Hence proved.

#### Question 42:

Prove the following trigonometric identities.

We need to prove

Solving the L.H.S, we get

= RHS

Hence proved.

#### Question 43:

Prove the following trigonometric identities.

We need to prove

Using the identity, we get

Further, using the property , we get

So,

Hence proved.

#### Question 44:

Prove the following trigonometric identities.

In the given question, we need to prove .

Here, we will first solve the LHS.

Now, using , we get

On further solving by taking the reciprocal of the denominator, we get,

Hence proved.

#### Question 45:

Prove the following trigonometric identities.

In the given question, we need to prove

Here, we will first solve the LHS.

Now, using, we get

On further solving by taking the reciprocal of the denominator, we get,

Now, taking common from both the numerator and the denominator, we get

Hence proved.

#### Question 46:

Prove the following trigonometric identities.

In the given question, we need to prove .

Using the property , we get

So,

$\frac{1+\mathrm{cos}\theta -{\mathrm{sin}}^{2}\theta }{\mathrm{sin}\theta \left(1+\mathrm{cos}\theta \right)}\phantom{\rule{0ex}{0ex}}=\frac{1+\mathrm{cos}\theta -\left(1-{\mathrm{cos}}^{2}\theta \right)}{\mathrm{sin}\theta \left(1+\mathrm{cos}\theta \right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cos}\theta +{\mathrm{cos}}^{2}\theta }{\mathrm{sin}\theta \left(1+\mathrm{cos}\theta \right)}\phantom{\rule{0ex}{0ex}}$

Solving further, we get

Hence proved.

#### Question 47:

Prove the following trigonometric identities.

(i)
(ii)
(iii)
(iv) $\left(\mathrm{sin}\theta +\mathrm{cos}\theta \right)\left(\mathrm{tan}\theta +\mathrm{cot}\theta \right)=\mathrm{sec}\theta +\mathrm{cosec}\theta$

(i) We have to prove the following identity-

Consider the LHS.

$\frac{1+\mathrm{cos}\theta +\mathrm{sin}\theta }{1+\mathrm{cos}\theta -\mathrm{sin}\theta }$

= RHS

Hence proved.

(ii) We have to prove the following identity-

Consider the LHS.

$\frac{\mathrm{sin}\theta -\mathrm{cos}\theta +1}{\mathrm{sin}\theta +\mathrm{cos}\theta -1}$

(Divide numerator and denominator by)

RHS

Hence proved.

(iii) We have to prove the following identity-

Consider the LHS.

= RHS

Hence proved.

(iv)
Consider the LHS.

= RHS
Hence proved.

#### Question 48:

Prove the following trigonometric identities.

In the given question, we need to prove .

Here, we will first solve the L.H.S.

Now, using, we get

On further solving, we get

Similarly we solve the R.H.S.

Now, using, we get

On further solving, we get

So, L.H.S = R.H.S

Hence proved.

#### Question 49:

Prove the following trigonometric identities.

tan2 A + cot2 A = sec2 A cosec2 A − 2

In the given question, we need to prove

Now, using and in L.H.S, we get

Further, using the identity, we get

Since L.H.S = R.H.S

Hence proved.

#### Question 50:

$\frac{\mathrm{tan}A}{1+\mathrm{sec}A}-\frac{\mathrm{tan}A}{1-\mathrm{sec}A}=2\mathrm{cosec}A$

Consider the LHS.
$\frac{\mathrm{tan}A}{1+\mathrm{sec}A}-\frac{\mathrm{tan}A}{1-\mathrm{sec}A}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{tan}A\left(1-\mathrm{sec}A\right)-\mathrm{tan}A\left(1+\mathrm{sec}A\right)}{\left(1+\mathrm{sec}A\right)\left(1-\mathrm{sec}A\right)}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{tan}A-\mathrm{tan}A\mathrm{sec}A-\mathrm{tan}A-\mathrm{tan}A\mathrm{sec}A}{\left(1-{\mathrm{sec}}^{2}A\right)}\phantom{\rule{0ex}{0ex}}=\frac{-2\mathrm{tan}A\mathrm{sec}A}{\left(1-{\mathrm{sec}}^{2}A\right)}\phantom{\rule{0ex}{0ex}}=\frac{-2\mathrm{tan}A\mathrm{sec}A}{-{\mathrm{tan}}^{2}A}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{sec}A}{\mathrm{tan}A}\phantom{\rule{0ex}{0ex}}=2\mathrm{cosec}A$
= RHS
Hence proved.

#### Question 51:

Prove the following trigonometric identities.

In the given question, we need to prove

Using and, we get

Further, using the property , we get

Hence proved.

#### Question 52:

Prove the following trigonometric identities.

In the given question, we need to prove

Using the identity, we get

Further, using the property , we get

Hence proved.

#### Question 53:

Prove the following trigonometric identities.

$\left(1+{\mathrm{tan}}^{2}A\right)+\left(1+\frac{1}{{\mathrm{tan}}^{2}A}\right)=\frac{1}{{\mathrm{sin}}^{2}A-{\mathrm{sin}}^{4}A}$

We need to prove .

Using the property , we get

Now, using , we get

Further, using the property, , we get

Hence proved.

#### Question 54:

Prove the following trigonometric identities.

sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B

We know that,

So have,

Hence proved.

#### Question 55:

Prove the following trigonometric identities.

(i)

(ii)

(i) We have to prove

Now,

Hence proved.

(ii) We have to prove

Now,

Hence proved.

#### Question 56:

Prove the following trigonometric identities.

cot2 A cosec2 B − cot2 B cosec2 A = cot2 A − cot2 B

We have to prove

We know that,

So,

Hence proved.

#### Question 57:

Prove the following trigonometric identities.

tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B

We have to prove

We know that,

So,

Hence proved.

#### Question 58:

Prove the following trigonometric identities.

If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2y2 = a2b2

Given that,

We have to prove

We know that,

So,

Hence proved.

#### Question 59:

Prove the following trigonometric identities.

If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ − 3 cos θ = ± 3.

Given:

We have to prove that 5sin θ – 3cos θ = ±3.

We know that,

Squaring the given equation, we have

⇒                                                   (5sin θ – 3cos θ)2 = 9
⇒                                                       5sin θ – 3cos θ = ±3

Hence proved.

#### Question 60:

Prove the following trigonometric identities.

If cosec θ + cot θ = m and cosec θ − cot θ = n, prove that mn = 1

Given:

We have to prove

We know that,

Multiplying the two equations, we have

Hence proved.

#### Question 61:

Prove the following trigonometric identities.

In the given question, we need to prove

Using the property and, we get

Taking the reciprocal of the denominator, we get

Further, using the identity, we get

Hence proved.

#### Question 62:

Prove the following trigonometric identities.

In the given question, we are given

We need to prove

Here L.H.S is

Now, solving the L.H.S, we get

Further using the property, we get

So,

Now, solving the R.H.S, we get

So,

Further using the property, we get,

So,

Hence proved.

#### Question 63:

Prove the following trigonometric identities.

In the given question, we need to prove

Now, using the identity in L.H.S, we get

Further using, we get

Also, from the identity , we get

Hence proved.

#### Question 64:

Prove the following trigonometric identities.

In the given question, we need to prove

Now, using and in L.H.S, we get

Further using the identity, we get

Further using the identity , we get

Now, from the identity, we get

So,

Hence proved.

#### Question 65:

Prove the following trigonometric identities.

(i)
(ii)  $\frac{1+\mathrm{sec}\theta -\mathrm{tan}\theta }{1+\mathrm{sec}\theta +\mathrm{tan}\theta }=\frac{1-\mathrm{sin}\theta }{\mathrm{cos}\theta }$

(i) In the given question, we need to prove

Taking common from the numerator and the denominator of the L.H.S, we get

Now, using the property , we get

Using , we get

Taking common from the numerator, we get

Using and , we get

Now, using the property , we get

Hence proved.

(ii)
Consider the LHS.

= RHS
Hence proved.

#### Question 66:

Prove the following trigonometric identities.

(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A

We have to prove

We know that,

So, we have

So, we have

Hence proved.

#### Question 67:

Prove the following trigonometric identities.

(1 + cot A − cosec A) (1 + tan A + sec A) = 2

We have to prove

We know that, .

So,

Hence proved.

#### Question 68:

Prove the following trigonometric identities.

(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)

We have to prove

We know that,

Consider the LHS.

Now, consider the RHS.

∴ LHS = RHS

Hence proved.

#### Question 69:

Prove the following trigonometric identities.

(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A

We have to prove

We know that,

So,

Hence proved.

#### Question 70:

Prove the following trigonometric identities.

(i)

(ii)

(i) We have to prove

So,

Hence proved.

(ii) We have to prove

We know that,

So,

Hence proved.

#### Question 71:

Prove the following trigonometric identities.

We have to prove

We know that,

So,

Hence proved.

#### Question 72:

Prove the following trigonometric identities.

We have to prove

We know that, .

So,

Hence proved.

#### Question 73:

Prove the following trigonometric identities.

sec4 A(1 − sin4 A) − 2 tan2 A = 1

We have to prove

We know that,

So,

Hence proved.

#### Question 74:

Prove the following trigonometric identities.

We have to prove .

We know that,

So,

Multiplying both the numerator and denominator by , we have

Hence proved.

#### Question 75:

Prove the following trigonometric identities.

We have prove that

We know that,

So,

Now,

Hence proved.

#### Question 76:

Prove the following trigonometric identities.

If

Given that,

We have to prove

We know that,

Squaring and then adding the above two equations, we have

#### Question 77:

Prove the following trigonometric identities.

If cosec θ − sin θ = a3, sec θ − cos θ = b3, prove that a2 b2 (a2 + b2) = 1

Given that,

We have to prove

We know that

Now from the first equation, we have

Again from the second equation, we have

Therefore, we have

Hence proved.

#### Question 78:

Prove the following trigonometric identities.

If a cos3 θ + 3 a cos θ sin2 θ = m, a sin3 θ + 3 a cos2 θ sin θ = n, prove that (m + n)2/3 + (mn)2/3 = 2a2/3

Given that,

We have to prove

Adding both the equations, we get

Also.

Therefore, we have

Hence proved.

#### Question 79:

Prove the following trigonometric identities.

If x = a cos3 θ, y = b sin3 θ, prove that ${\left(\frac{x}{a}\right)}^{2/3}+{\left(\frac{y}{b}\right)}^{2/3}=1$

Given:

We have to prove

We know that

So, we have

Hence proved.

#### Question 80:

Prove the following trigonometric identities.

If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove that a2 + b2 = m2 + n2

Given:

We have to prove

We know that,

Now, squaring and adding the two equations, we get

Hence proved.

#### Question 81:

Prove the following trigonometric identities.

if cos A + cos2 A = 1, prove that sin2 A + sin4 A = 1

Given:

We have to prove

Now,

Therefore, we have

Hence proved.

#### Question 82:

If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1

Given:

We have to prove

From the given equation, we have

Therefore, we have

Hence proved.

#### Question 83:

Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)

Show that one of the values of each member of this equality is sin α sin β sin γ

Given:

Let us assume that

We know that,

Then, we have

Therefore, we have

Taking the expression with the positive sign, we have

#### Question 84:

If sin θ + cos θ = x, prove that .

Given:

Squaring the given equation, we have

Squaring the last equation, we have

Therefore, we have

Hence proved.

#### Question 85:

If x = a sec θ cos Ď•, y = b sec θ sin Ď• and z = c tan θ, show that $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}-\frac{{z}^{2}}{{c}^{2}}=1$.

Given:

We have to prove that .

Squaring the above equations and then subtracting the third from the sum of the first two, we have

Hence proved.

#### Question 86:

If $\mathrm{sin}\theta +2\mathrm{cos}\theta =1$ prove that $2\mathrm{sin}\theta -\mathrm{cos}\theta =2$.

It is given that,

#### Question 1:

If , find all other trigonometric ratios of angle θ.

Given:

Now, we have to find all the other trigonometric ratios.

We have the following right angle triangle.

From the above figure,

$\mathrm{Perpendicular}=\sqrt{{\mathrm{Hypotenuse}}^{2}-{\mathrm{Base}}^{2}}$

Therefore,

#### Question 2:

If , find all other trigonometric ratios of angle θ.

Given:

We have to find all the trigonometric ratios.

We have the following right angle triangle.

From the above figure,

$\mathrm{Base}=\sqrt{{\mathrm{Hypotenuse}}^{2}-{\mathrm{Perpendicular}}^{2}}\phantom{\rule{0ex}{0ex}}⇒BC=\sqrt{A{C}^{2}-A{B}^{2}}\phantom{\rule{0ex}{0ex}}⇒BC=\sqrt{{\left(\sqrt{2}\right)}^{2}-{1}^{2}}\phantom{\rule{0ex}{0ex}}⇒BC=1$

#### Question 3:

If , find the value of .

Given:
We have to find the value of the expression

We know that,

Therefore, the given expression can be written as

Hence, the value of the given expression is .

#### Question 4:

If 4tanθ = 3, evaluate

Given: 4tanθ = 3 ⇒ tan θ = $\frac{3}{4}$
Let us suppose a right angle triangle ABC right angled at B, with one of the acute angle θ. Let the sides be BC = 3and AB = 4k, where k is a positive number.

By Pythagoras theorem, we get
${\mathrm{AC}}^{2}={\mathrm{BC}}^{2}+{\mathrm{AB}}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{AC}}^{2}={\left(3k\right)}^{2}+{\left(4k\right)}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{AC}}^{2}=9{k}^{2}+16{k}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{AC}=\sqrt{25{k}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{AC}=±5k\phantom{\rule{0ex}{0ex}}$
Ignoring AC = − 5k , as k is a positive number, we get
AC = 5k
If $\mathrm{tan}\theta =\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3}{4}$, then $\mathrm{sin}\theta =\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{5}$ and $\mathrm{cos}\theta =\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4}{5}$
Putting the values in $\left(\frac{4\mathrm{sin}\theta -\mathrm{cos}\theta +1}{4\mathrm{sin}\theta +\mathrm{cos}\theta -1}\right)$, we get
$\left(\frac{4×\frac{3}{5}-\frac{4}{5}+1}{4×\frac{3}{5}+\frac{4}{5}-1}\right)=\left(\frac{\frac{12-4+5}{5}}{\frac{12+4-5}{5}}\right)=\frac{13}{11}$

#### Question 5:

If , find the value of .

Given:
We have to find the value of the expression .

From the above figure, we have

Therefore,

Hence, the value of the given expression is 25.

#### Question 6:

If , find the value of .

Given:
We have to find the value of the expression

We know that,

Using the identity , we have

Hence, the value of the given expression is $\frac{3}{5}$.

#### Question 7:

If , find the value of .

Given:
We have to find the value of the expression

We know that,

Therefore,

Hence, the value of the given expression is 2.

#### Question 8:

If , find the value of .

Given:

We have to find the value of the expression .

We know that,

Therefore,

Hence, the value of the given expression is $\frac{21}{8}$.

#### Question 9:

If 3cosθ = 1, find the value of

Given:

We have to find the value of the expression .

We have,

Therefore,

Hence, the value of the expression is 10.

#### Question 10:

If , find the value of sin2θ − cos2θ.

Given:

We have to find the value of .

Therefore,

Hence, the value of the expression is $\frac{1}{3}$.

#### Question 11:

If , find the value of .

Given:

We have to find the value of the expression .

Now,

Therefore,

Hence, the value of the expression is 3.

#### Question 12:

If sin θ + cos θ = , find cot θ.

Given:

We have to find the value of .

Now,

Hence,

#### Question 13:

If   , then find the value of $\mathrm{\theta }$ .

It is given that,

#### Question 1:

If sec θ + tan θ = x, then sec θ =

(a) $\frac{{x}^{2}+1}{x}$
(b) $\frac{{x}^{2}+1}{2x}$
(c) $\frac{{x}^{2}-1}{2x}$
(d) $\frac{{x}^{2}-1}{x}$

Given:

We know that,

Now,

Adding the two equations, we get

Therefore, the correct choice is (b).

#### Question 2:

If $\mathrm{sec\theta }+\mathrm{tan\theta }=x$, then $\mathrm{tan\theta }=$

(a) $\frac{{x}^{2}+1}{x}$
(b) $\frac{{x}^{2}-1}{x}$
(c) $\frac{{x}^{2}+1}{2x}$
(d) $\frac{{x}^{2}-1}{2x}$

Given:

We know that,

Now,

Subtracting the second equation from the first equation, we get

Therefore, the correct choice is (d).

#### Question 3:

is equal to

(a) sec θ + tan θ
(b) sec θ − tan θ
(c) sec2 θ + tan2 θ
(d) sec2 θ − tan2 θ

The given expression is .

Multiplying both the numerator and denominator under the root by , we have

Therefore, the correct option is (a).

#### Question 4:

The value of is

(a) cot θ − cosec θ
(b) cosec θ + cot θ
(c) cosec2 θ + cot2 θ
(d) (cot θ + cosec θ)2

The given expression is .

Multiplying both the numerator and denominator under the root by, we have

Therefore, the correct choice is (b).

#### Question 5:

sec4 A − sec2 A is equal to

(a) tan2 A − tan4 A
(b) tan4 A − tan2 A
(c) tan4 A + tan2 A
(d) tan2 A + tan4 A

The given expression is .

Taking common from both the terms, we have

Disclaimer: The options given in (c) and (d) are same by the commutative property of addition.

Therefore, the correct options are (c) or (d).

#### Question 6:

cos4 A − sin4 A is equal to

(a) 2 cos2 A + 1
(b) 2 cos2 A − 1
(c) 2 sin2 A − 1
(d) 2 sin2 A + 1

The given expression is .

Factorising the given expression, we have

Therefore, the correct option is (b).

#### Question 7:

is equal to

(a)
(b)
(c)
(d)

The given expression is .

Multiplying both the numerator and denominator under the root by , we have

Therefore, the correct option is (c).

#### Question 8:

is equal to

(a) 0
(b) 1
(c) sin θ + cos θ
(d) sin θ − cos θ

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (c).

#### Question 9:

The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is

(a) 1
(b) 2
(c) 4
(d) 0

The given expression is

Simplifying the given expression, we have

Therefore, the correct option is (b).

#### Question 10:

is equal to

(a) 2 tan θ
(b) 2 sec θ
(c) 2 cosec θ
(d) 2 tan θ sec θ

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (c).

#### Question 11:

(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to

(a) 0
(b) 1
(c) −1
(d) None of these

The given expression is

Simplifying the given expression, we have

Therefore, the correct option is (b).

#### Question 12:

If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =

(a) a2 b2
(b) ab
(c) a4 b4
(d) a2 + b2

Given:

So,

We know that,

Therefore,

Hence, the correct option is (a).

#### Question 13:

If x = a sec θ and y = b tan θ, then b2x2a2y2 =

(a) ab
(b) a2b2
(c) a2 + b2
(d) a2 b2

Given:

So,

We know that,

Therefore,

Hence, the correct option is (d).

#### Question 14:

is equal to

(a) 0
(b) 1
(c) −1
(d) 2

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (b).

#### Question 15:

2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to

(a) 0
(b) 1
(c) −1
(d) None of these

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (c).

#### Question 16:

If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =

(a) 7
(b) 12
(c) 25
(d) None of these

Given:

Squaring and then adding the above two equations, we have

Hence, the correct option is (c).

#### Question 17:

If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2q2 =

(a) a2 b2
(b) b2a2
(c) a2 + b2
(d) ba

Given:

Squaring both the equations and then subtracting the second from the first, we have

Hence, the correct option is (b).

#### Question 18:

The value of sin2 29° + sin2 61° is

(a) 1
(b) 0
(c) 2 sin2 29°
(d) 2 cos2 61°

The given expression is .

Hence, the correct option is (a).

#### Question 19:

If x = r sin θ cos Ď•, y = r sin θ sin Ď• and z = r cos θ, then

(a) ${x}^{2}+{y}^{2}+{z}^{2}={r}^{2}$
(b) ${x}^{2}+{y}^{2}-{z}^{2}={r}^{2}$
(c) ${x}^{2}-{y}^{2}+{z}^{2}={r}^{2}$
(d) ${z}^{2}+{y}^{2}-{x}^{2}={r}^{2}$

Given:

Squaring and adding these equations, we get

Hence, the correct option is (a).

#### Question 20:

If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =

(a) −1
(b) 1
(c) 0
(d) None of these

Given:

Now,

Hence, the correct option is (b).

#### Question 21:

If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =

(a) m2n2
(b) m2n2
(c) n2m2
(d) m2 + n2

Given:

Squaring and adding these equations, we have

Hence, the correct option is (d).

#### Question 22:

If cos A + cos2 A = 1, then sin2 A + sin4 A =

(a) −1
(b) 0
(c) 1
(d) None of these

Given:

So,

Hence, the correct option is (c).

#### Question 23:

If x = a sec θ cos Ď•, y = b sec θ sin Ď• and z = c tan θ, then $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}$=

(a) $\frac{{z}^{2}}{{c}^{2}}$
(b) $1-\frac{{z}^{2}}{{c}^{2}}$
(c) $\frac{{z}^{2}}{{c}^{2}}-1$
(d) $1+\frac{{z}^{2}}{{c}^{2}}$

Given:

Now,

Hence, the correct option is (d).

#### Question 24:

If a cos θ − b sin θ = c, then a sin θ + b cos θ =

(a) $±\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}$
(b)
(c) $±\sqrt{{c}^{2}-{a}^{2}-{b}^{2}}$
(d) None of these

Given:

Squaring on both sides, we have

Hence, the correct option is (b).

#### Question 25:

9 sec2 A − 9 tan2 A is equal to

(a) 1
(b) 9
(c) 8
(d) 0

Given:

We know that,

Therefore,

Hence, the correct option is (b).

#### Question 26:

(1 + tan θ + sec θ) (1 + cot θ − cosec θ) =

(a) 0
(b) 1
(c) 1
(d) −1

The given expression is .

Simplifying the given expression, we have

Disclaimer: None of the given options match with the answer.

#### Question 27:

(sec A + tan A) (1 − sin A) =

(a) sec A
(b) sin A
(c) cosec A
(d) cos A

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (d).

#### Question 28:

is equal to

(a) sec2 A
(b) −1
(c) cot2 A
(d) tan2 A

Given:

Therefore, the correct option is (d).

#### Question 29:

If sin  , then the value of  is
(a) 1     (b)  $-1$      (c)   $\frac{1}{2}$      (d)  $\frac{1}{4}$

It is given that,
$\mathrm{sin}\theta -\mathrm{cos}\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\theta =1\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\theta =\mathrm{tan}45°\phantom{\rule{0ex}{0ex}}⇒\theta =45°$
$\therefore {\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta \phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{4}45°+{\mathrm{cos}}^{4}45°\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{\sqrt{2}}\right)}^{4}+{\left(\frac{1}{\sqrt{2}}\right)}^{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}+\frac{1}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$
Hence, the correct answer is option (c).

#### Question 30:

The value of sin ( is equal to
(a) 2 cos $\mathrm{\theta }$    (b) 0     (c)     2 sin $\mathrm{\theta }$    (d) 1

We know that, $\mathrm{sin}\left(90-\theta \right)=\mathrm{cos}\theta$.
So, $\mathrm{sin}\left(45°+\theta \right)=\mathrm{cos}\left[90-\left(45°+\theta \right)\right]=\mathrm{cos}\left(45°-\theta \right)$
$\therefore \mathrm{sin}\left(45°+\mathrm{\theta }\right)-\mathrm{cos}\left(45°-\mathrm{\theta }\right)\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\left(45°-\mathrm{\theta }\right)-\mathrm{cos}\left(45°-\mathrm{\theta }\right)\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Hence, the correct answer is option (b).

#### Question 31:

If $∆ABC$ is right angled at C , then the value of cos (  ) is

In a right angled triangle ABC, $\angle \mathrm{C}$ is a right angle.
We know that, the sum of angles of a triangle is 180º.
$\therefore \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{A}+\angle \mathrm{B}+90°=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{A}+\angle \mathrm{B}=90°$
$\therefore \mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{cos}90°=0$
Hence, the correct answer is option (a).

#### Question 32:

If cos $9\mathrm{\theta }$ = sin $\mathrm{\theta }$ and  $9\mathrm{\theta }$  < 900 , then the value of tan  is

It is given that,

Disclaimer: Answer of the given question is not matching with the options provided in the textbook.

#### Question 33:

If  cos () = 0 , then sin  can be reduced to

It is given that,

Hence, the correct answer is option (b).

#### Question 1:

If cosec θ + cot θ = 3, then cos θ = _________.

$\mathrm{cosec}\theta +\mathrm{cot}\theta =3\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sin}\theta }+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }=3\phantom{\rule{0ex}{0ex}}⇒\frac{1+\mathrm{cos}\theta }{\mathrm{sin}\theta }=3$
$⇒1+\mathrm{cos}\theta =3\mathrm{sin}\theta$
Squaring on both sides, we get
$1+{\mathrm{cos}}^{2}\theta +2\mathrm{cos}\theta =9\left(1-{\mathrm{cos}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}⇒10{\mathrm{cos}}^{2}\theta +2\mathrm{cos}\theta -8=0\phantom{\rule{0ex}{0ex}}⇒5{\mathrm{cos}}^{2}\theta +\mathrm{cos}\theta -4=0\phantom{\rule{0ex}{0ex}}⇒5{\mathrm{cos}}^{2}\theta +5\mathrm{cos}\theta -4\mathrm{cos}\theta -4=0$

If cosec θ + cot θ = 3, then cos θ = .

#### Question 2:

If cosec θ – cot θ = 2, then find the value of cosec2θ + cot2θ is ______.

We know that,

Now,
${\left(\mathrm{cosec}\theta -\mathrm{cot}\theta \right)}^{2}+{\left(\mathrm{cosec}\theta +\mathrm{cot}\theta \right)}^{2}={2}^{2}+{\left(\frac{1}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cosec}}^{2}\theta +{\mathrm{cot}}^{2}\theta -2\mathrm{cosec}\theta \mathrm{cot}\theta +{\mathrm{cosec}}^{2}\theta +{\mathrm{cot}}^{2}\theta +2\mathrm{cosec}\theta \mathrm{cot}\theta =4+\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒2\left({\mathrm{cosec}}^{2}\theta +{\mathrm{cot}}^{2}\theta \right)=\frac{17}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cosec}}^{2}\theta +{\mathrm{cot}}^{2}\theta =\frac{17}{8}\phantom{\rule{0ex}{0ex}}$

If cosec θ – cot θ = 2, then find the value of cosec2θ + cot2θ is .

#### Question 3:

The value of sin2 20° + sin2 70° is __________.

We know that,

The value of sin2 20° + sin2 70° is _____1_____.

#### Question 4:

The value of sin θ cos(90° – θ) + cos θ sin (90° – θ) is ___________.

The value of sin θ cos(90° – θ) + cos θ sin (90° – θ) is _____1_____.

#### Question 5:

The value of $\frac{{\mathrm{sin}}^{4}\mathrm{\theta }-{\mathrm{cos}}^{4}\mathrm{\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }-{\mathrm{cos}}^{2}\mathrm{\theta }}$ is ____________.

$={\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \phantom{\rule{0ex}{0ex}}=1$

The value of $\frac{{\mathrm{sin}}^{4}\mathrm{\theta }-{\mathrm{cos}}^{4}\mathrm{\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }-{\mathrm{cos}}^{2}\mathrm{\theta }}$ is ______1______.

#### Question 6:

If sec θ + tan θ = m, then the value of sec4 θ – tan4 θ – 2 sec θ tan θ is _________.

$={\left[\frac{\left(\mathrm{sec}\theta +\mathrm{tan}\theta \right)\left(\mathrm{sec}\theta -\mathrm{tan}\theta \right)}{\mathrm{sec}\theta +\mathrm{tan}\theta }\right]}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{{\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta }{\mathrm{sec}\theta +\mathrm{tan}\theta }\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{m}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{{m}^{2}}$

If sec θ + tan θ = m, then the value of sec4 θ – tan4 θ – 2 sec θ tan θ is .

#### Question 7:

If sin4 A – cos4 A = 1 and 0 < A ≤ 90°, then A = __________.

$⇒2{\mathrm{cos}}^{2}A=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}A=0\phantom{\rule{0ex}{0ex}}⇒A=90°$

If sin4 A – cos4 A = 1 and 0 < A ≤ 90°, then A = .

#### Question 8:

The value of is __________.

$\frac{{\mathrm{cos}}^{4}\theta +{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta +{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta }\phantom{\rule{0ex}{0ex}}=\frac{{\left({\mathrm{cos}}^{2}\theta \right)}^{2}+{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta +{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta }\phantom{\rule{0ex}{0ex}}=\frac{{\left(1-{\mathrm{sin}}^{2}\theta \right)}^{2}+{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta +{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1+{\mathrm{sin}}^{4}\theta -2{\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta +{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta }$
$=\frac{1-{\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta +{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta }\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{cos}}^{2}\theta +{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta }{{\mathrm{cos}}^{2}\theta +{\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta }\phantom{\rule{0ex}{0ex}}=1$

The value of $\frac{{\mathrm{cos}}^{4}\mathrm{\theta }+{\mathrm{cos}}^{2}{\mathrm{\theta sin}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }}{{\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}{\mathrm{\theta sin}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{4}\mathrm{\theta }}$ is _____1_____.

#### Question 9:

If 2 sin θ + 3 cos θ = 2, then 3 sin θ – 2 cos θ = _________.

If 2 sin θ + 3 cos θ = 2, then 3 sin θ – 2 cos θ = ____±3_____.

#### Question 10:

If sin θ – cos θ = $\frac{3}{5}$, then sin θ cos θ = _______.

$\mathrm{sin}\theta -\mathrm{cos}\theta =\frac{3}{5}$
Squaring on both sides, we get
${\left(\mathrm{sin}\theta -\mathrm{cos}\theta \right)}^{2}={\left(\frac{3}{5}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta -2\mathrm{sin}\theta \mathrm{cos}\theta =\frac{9}{25}\phantom{\rule{0ex}{0ex}}⇒1-2\mathrm{sin}\theta \mathrm{cos}\theta =\frac{9}{25}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sin}\theta \mathrm{cos}\theta =1-\frac{9}{25}=\frac{16}{25}\phantom{\rule{0ex}{0ex}}$
$⇒\mathrm{sin}\theta \mathrm{cos}\theta =\frac{8}{25}$

If sin θ – cos θ = $\frac{3}{5}$, then sin θ cos θ = .

#### Question 11:

The value of sin22° + sin24° + sin26° + ... + sin290° is ____________.

sin22° + sin24° + sin26° + ... + sin290°

= sin22° + sin24° + sin26° + ... + sin284° + sin286° + sin288°+ sin290°

= [(sin22° + sin288°) + (sin24° + sin286°) + (sin26° + sin284°) + .... 22 pair of terms] + sin290°

= [(sin22° + cos22°) + (sin24° + cos24°) + (sin26° + cos26°) + .... 22 pair of terms] + sin290°                 [sin(90º − θ) = cosθ]

= (1 + 1 + 1 + ... 22 terms) + 1                              [sin2θ + cos2θ = 1 and sin90° = 1]

= 22 + 1

= 23

The value of sin22° + sin24° + sin26° + ... + sin290° is ______23______.

#### Question 12:

If

$\frac{{\mathrm{sin}}^{2}\theta -3\mathrm{sin}\theta +2}{{\mathrm{cos}}^{2}\theta }=1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\theta -3\mathrm{sin}\theta +2=1-{\mathrm{sin}}^{2}\theta \phantom{\rule{0ex}{0ex}}⇒2{\mathrm{sin}}^{2}\theta -3\mathrm{sin}\theta +1=0$

Here, $\theta$ cannot take the value 90º. For $\theta =90°$, the LHS of the given equation is not defined.

Thus, the value of $\theta$ is $30°$.

If

#### Question 13:

If cos θ + cos2θ = 1, then sin2θ + sin4θ = _________.

$\mathrm{cos}\theta +{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =1-{\mathrm{cos}}^{2}\theta \phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta ={\mathrm{sin}}^{2}\theta$
Squaring on both sides, we get
${\mathrm{cos}}^{2}\theta ={\mathrm{sin}}^{4}\theta \phantom{\rule{0ex}{0ex}}⇒1-{\mathrm{sin}}^{2}\theta ={\mathrm{sin}}^{4}\theta \phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta =1$

If cos θ + cos2θ = 1, then sin2θ + sin4θ = ____1_____.

#### Question 14:

If

Comparing with the given expression, we get

A = 1 and B = 1

A + B = 2

If

#### Question 15:

The value of (cosec θ – sin θ)  (secθ – cos θ) (tan θ + cot θ) is _________.

$\left(\mathrm{cosec}\theta -\mathrm{sin}\theta \right)\left(\mathrm{sec}\theta -\mathrm{cos}\theta \right)\left(\mathrm{tan}\theta +\mathrm{cot}\theta \right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{\mathrm{sin}\theta }-\mathrm{sin}\theta \right)\left(\frac{1}{\mathrm{cos}\theta }-\mathrm{cos}\theta \right)\left(\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1-{\mathrm{sin}}^{2}\theta }{\mathrm{sin}\theta }\right)\left(\frac{1-{\mathrm{cos}}^{2}\theta }{\mathrm{cos}\theta }\right)\left(\frac{{\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta }{\mathrm{sin}\theta \mathrm{cos}\theta }\right)$

The value of (cosec θ – sin θ)  (secθ – cos θ) (tan θ + cot θ) is ___1___.

#### Question 16:

If , then the value of k is _________.

$\frac{1}{1+\mathrm{sin}\theta }+\frac{1}{1-\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1-\mathrm{sin}\theta +1+\mathrm{sin}\theta }{\left(1+\mathrm{sin}\theta \right)\left(1-\mathrm{sin}\theta \right)}\phantom{\rule{0ex}{0ex}}=\frac{2}{1-{\mathrm{sin}}^{2}\theta }$

Comparing with the given expression, we get

k = 2

If , then the value of k is ____2____.

#### Question 17:

$\sqrt{-4+\sqrt{8+16{\mathrm{cosec}}^{4}\theta +{\mathrm{sin}}^{4}\theta }}\phantom{\rule{0ex}{0ex}}=\sqrt{-4+\sqrt{\left[{\left(4{\mathrm{cosec}}^{2}\theta \right)}^{2}+2×4{\mathrm{cosec}}^{2}\theta ×{\mathrm{sin}}^{2}\theta +{\left({\mathrm{sin}}^{2}\theta \right)}^{2}\right]}}\phantom{\rule{0ex}{0ex}}=\sqrt{-4+\sqrt{{\left(4{\mathrm{cosec}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)}^{2}}}$
$=\sqrt{-4+4{\mathrm{cosec}}^{2}\theta +{\mathrm{sin}}^{2}\theta }\phantom{\rule{0ex}{0ex}}=\sqrt{\left[{\left(2\mathrm{cosec}\theta \right)}^{2}-2×2\mathrm{cosec}\theta ×\mathrm{sin}\theta +{\mathrm{sin}}^{2}\theta \right]}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(2\mathrm{cosec}\theta -\mathrm{sin}\theta \right)}^{2}}\phantom{\rule{0ex}{0ex}}=2\mathrm{cosec}\theta -\mathrm{sin}\theta$

Comparing with the given expression, we get

A = 2 and B = −1

#### Question 18:

If tan θ and 0 < θ < 90°, then k = ________.

Comparing with the given expression, we get

k = 1

If tan θ and 0 < θ < 90°, then k = ____1____.

#### Question 19:

If (tan θ + 2) (2tan θ + 1) = A tan θ + B sec2θ, then AB = _________.

Comparing with the given expression, we get

A = 5 and B = 2

AB = 5 × 2 = 10

If (tan θ + 2) (2tan θ + 1) = A tan θ + B sec2θ, then AB = ____10____.

#### Question 1:

Define an identity.

An identity is an equation which is true for all values of the variable (s).

For example,

Any number of variables may involve in an identity.

An example of an identity containing two variables is

The above are all about algebraic identities. Now, we define the trigonometric identities.

An equation involving trigonometric ratios of an angle (say) is said to be a trigonometric identity if it is satisfied for all valued of for which the trigonometric ratios are defined.

For examples,

${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\phantom{\rule{0ex}{0ex}}1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta \phantom{\rule{0ex}{0ex}}1+{\mathrm{cot}}^{2}\theta ={\mathrm{cosec}}^{2}\theta$

#### Question 2:

What is the value of (1 − cos2 θ) cosec2 θ?

We have,

#### Question 3:

What is the value of (1 + cot2 θ) sin2 θ?

We have,

#### Question 4:

What is the value of ?

We have,

#### Question 5:

If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.

Given:

We know that,

Therefore,

Hence,

#### Question 6:

If cosec θ − cot θ = α, write the value of cosec θ + cot α.

Given:

We know that,

Therefore,

Hence,

#### Question 7:

Write the value of cosec2 (90° − θ) − tan2 θ.

We have,

We know that,

Therefore, ${\mathrm{cosec}}^{2}\left(90°-\theta \right)-{\mathrm{tan}}^{2}\theta =1$

#### Question 8:

Write the value of sin A cos (90° − A) + cos A sin (90° − A).

We have,

We know that,

Therefore, $\mathrm{sin}A\mathrm{cos}\left(90°-A\right)+\mathrm{cos}A\mathrm{sin}\left(90°-A\right)=1$

#### Question 9:

Write the value of .

We have,

We know that,

Therefore, ${\mathrm{cot}}^{2}\mathrm{\theta }-\frac{1}{{\mathrm{sin}}^{2}\mathrm{\theta }}=-1$

#### Question 10:

If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?

Given:

x = a sinθ and y = b cosθ

So,

${b}^{2}{x}^{2}+{a}^{2}{y}^{2}={b}^{2}{\left(a\mathrm{sin\theta }\right)}^{2}+{a}^{2}{\left(b\mathrm{cos\theta }\right)}^{2}\phantom{\rule{0ex}{0ex}}={a}^{2}{b}^{2}{\mathrm{sin}}^{2}\mathrm{\theta }+{a}^{2}{b}^{2}{\mathrm{cos}}^{2}\mathrm{\theta }\phantom{\rule{0ex}{0ex}}={a}^{2}{b}^{2}\left({\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }\right)$

We know that,

Therefore, ${b}^{2}{x}^{2}+{a}^{2}{y}^{2}={a}^{2}{b}^{2}$

#### Question 11:

If , what is the value of cotθ + cosecθ?

Given:

We know that,

We have,

Hence, the value of cotθ + cosecθ is 2.

#### Question 12:

What is the value of 9cot2 θ − 9cosec2 θ?

We have,

We know that,

Therefore, $9{\mathrm{cot}}^{2}\mathrm{\theta }-9{\mathrm{cosec}}^{2}\mathrm{\theta }=-9$

#### Question 13:

What is the value of .

We have,

We know that,

Therefore,

#### Question 14:

What is the value of .

We have,

We know that,

Therefore,

#### Question 15:

What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?

We have,

We know that,

Therefore,

#### Question 16:

If , find the value of tan A + cot A.

Given:

We know that,

Therefore,

#### Question 17:

If , then find the value of 2cot2 θ + 2.

Given:

We know that,

Therefore,

#### Question 18:

If , then find the value of 9tan2 θ + 9.

Given:

We know that,

Therefore,

#### Question 19:

If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.

Given:

Hence, the value of k is 1.

#### Question 20:

If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.

Given:

Thus, the value of λ is 1.

#### Question 21:

If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.

Given:

$⇒\lambda =1×1=1$

Hence, the value of λ is 1.

#### Question 22:

If 5x = sec θ and , find the value of $5\left({x}^{2}-\frac{1}{{x}^{2}}\right)$.

Given:

We know that,

#### Question 23:

If cosec θ = 2x and , find the value of $2\left({x}^{2}-\frac{1}{{x}^{2}}\right)$.

Given:

We know that,

#### Question 24:

Write True' or False' and justify your answer in each of the following :
(i) The value of  is , where 'x'  is a positive real number .
(ii) , where a and b are two distinct numbers such that ab > 0.

(iii) The value of  is positive .

(iv) The value of the expression  is negative .

(v) The value of   is always greater than 1 .

(i)

(ii)
It is given that,

(iii)
${\mathrm{cos}}^{2}23°-{\mathrm{sin}}^{2}67°\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{2}\left(90°-23°\right)-{\mathrm{sin}}^{2}67°\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{2}67°-{\mathrm{sin}}^{2}67°\phantom{\rule{0ex}{0ex}}=0$
Which is not positive, the given statement is false.

(iv)
Consider the table.

 $\theta$ 0º 30º 45º 60º 90º sin$\theta$ 0 $\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ 1 cos$\theta$ 1 $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{2}$ 0

Here,

Therefore, the given statement is false.

(v)
Consider the table.
 $\theta$ 0º 30º 45º 60º 90º sin$\theta$ 0 $\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ 1 cos$\theta$ 1 $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{2}$ 0
Here,

#### Question 25:

What is the value of cos2 67° – sin2 23°?