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#### Question 27:

If $3x=\mathrm{cosec}\theta$ and $\frac{3}{x}=\mathrm{cot}\theta$, find the value of $3\left({x}^{2}-\frac{1}{{x}^{2}}\right)$.             [CBSE 2010]

$3\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{9}{3}\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(9{x}^{2}-\frac{9}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[{\left(3x\right)}^{2}-{\left(\frac{3}{x}\right)}^{2}\right]$
$=\frac{1}{3}\left[{\left(\mathrm{cosec}\theta \right)}^{2}-{\left(\mathrm{cot}\theta \right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left({\mathrm{cosec}}^{2}\theta -{\mathrm{cot}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(1\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$

#### Question 28:

In the adjoining figure, ∆ABC is a right-angled triangle in which ∠B = 90°, ∠A = 30° and AC = 20 cm.
Find (i) BC, (ii) AB.

From the given right-angled triangle, we have:

#### Question 29:

In the adjoining figure, ∆ABC is right-angled at B and ∠A = 30°. If BC = 6 cm, find (i) AB, (ii) AC.

From the given right-angled triangle, we have:

#### Question 30:

In the adjoining figure, ∆ABC is a right-angled at B and ∠A = 45°. If AC$3\sqrt{2}$ cm,
find (i) BC, (ii) AB.

From right-angled ∆ABC, we have:

#### Question 1:

sin 60° cos 30° + cos 60° sin 30°

On substituting the values of various T-ratios, we get:
sin 60o cos 30o + cos 60o sin 30o
$=\left(\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}+\frac{1}{2}×\frac{1}{2}\right)=\left(\frac{3}{4}+\frac{1}{4}\right)=\frac{4}{4}=1$

#### Question 2:

sin 60° cos 30° − cos 60° sin 30°

Ans

#### Question 3:

cos 60° cos 30° − sin 60° sin 30°

On substituting the values of various T-ratios, we get:
cos 60o cos 30o − sin 60o sin 30o
$=\left(\frac{1}{2}×\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}×\frac{1}{2}\right)=\left(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\right)=0$

#### Question 4:

cos 45° cos 30° + sin 45° sin 30°

On substituting the values of various T-ratios, we get:
cos 45o cos 30o + sin 45o  sin 30o

#### Question 5:

tan 30° cosec 60° + tan 60° sec 30°

As we know that,

#### Question 6:

2 cos2 60° + 3 sin2 45° − 3 sin2 30° + 2 cos2 90°

On substituting the values of various T-ratios, we get:
2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o

As we know that,

#### Question 8:

cot230° − 2cos230° − $\frac{3}{4}$ sec245° + $\frac{1}{4}$ cosec230°

On substituting the values of various T-ratios, we get:
cot2 30o − 2 cos2 30o − $\frac{3}{4}$ sec2 45o + $\frac{1}{4}$ cosec2 30o

#### Question 9:

(cos 0° + cos 45° + sin 30°) (sin 90° – cos 45° + cos 60°)

As we know that,

#### Question 10:

Show that:
(i) $\frac{1-\mathrm{sin}60°}{\mathrm{cos}60°}=\frac{\mathrm{tan}60°-1}{\mathrm{tan}60°+1}$
(ii) $\frac{\mathrm{cos}30°+\mathrm{sin}60°}{1+\mathrm{sin}30°+\mathrm{cos}60°}=\mathrm{cos}30°$

(i)

Hence, LHS = RHS

(ii)

Hence, LHS = RHS

1sin60°cos60°

#### Question 11:

Verify each of the following:
(i) sin 60° cos 30° − cos 60° sin 30° = sin 30°
(ii) cos 60° cos 30° + sin 60° sin 30° = cos 30°
(iii) 2 sin 30° cos 30° = sin 60°
(iv) 2 sin 45° cos 45° = sin 90°

(i) sin 60o cos 30o − cos 60o sin 30o

∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o

(ii) cos 60o cos 30o + sin 60o sin 30o

∴ cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) 2 sin 30o cos 30o

∴ 2 sin 30o cos 30o = sin 60o

(iv) 2 sin 45o cos 45o
$=2×\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}=1$
Also, sin 90o = 1
∴ 2 sin 45o cos 45o = sin 90o

#### Question 12:

If A = 45°, verify that:
(i) sin 2A = 2 sin A cos A
(ii) cos 2A = 2 cos2 A − 1 = 1 − 2 sin2 A

A = 45o
⇒ 2A = 2 $×$ 45o = 90o

(i) sin 2A = sin 90o = 1
2 sin A cos A = 2 sin 45o cos 45o
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90o = 0
2 cos2 A − 1 = 2 cos2 45o − 1 =
Now, 1 − 2 sin2 A
∴ cos 2A 2 cos2 A − 1 = 1 − 2 sin2 A

#### Question 13:

If A = 30°, verify that:
(i) $\mathrm{sin}2A=\frac{2\mathrm{tan}A}{1+{\mathrm{tan}}^{2}A}$
(ii) $\mathrm{cos}2A=\frac{1-{\mathrm{tan}}^{2}A}{1+{\mathrm{tan}}^{2}A}$
(iii) $\mathrm{tan}2A=\frac{2\mathrm{tan}A}{1-{\mathrm{tan}}^{2}A}$

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

​(i) sin 2A = sin 60o = $\frac{\sqrt{3}}{2}$

(ii) cos 2A = cos 60o = $\frac{1}{2}$

(iii) tan 2A = tan 60o = $\sqrt{3}$

=2tanA1+tan2A

#### Question 14:

If A = 60° and B = 30°, verify that:
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B − sin A sin B

A = 60o and B = 30o
Now, A + B = 60o + 30o​ = 90o
Also, A − B = 60o − 30o = 30o

(i) sin (A + B) = sin 90o = 1
sin A cos B + cos A sin B = sin 60o cos 30o + cos 60o sin 30o

∴ sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos 90o = 0
cos A cos B − sin A sin B =cos 60o cos 30o − sin 60o sin 30o
∴​ cos (A + B) = cos A cos B − sin A sin B

#### Question 15:

If A = 60° and B = 30°, verify that:

(i) sin (A B) = sin A cos B − cos A sin B
(ii) cos (AB) = cos A cos B + sin A sin B
(iii) tan (A − B) =

(i) sin (A − B) = sin 30o = $\frac{1}{2}$
sin A cos B − cos A sin B = sin 60o cos 30o − cos 60o sin 30o

∴ sin (A − B) = sin A cos B − cos A sin B

(ii) cos (AB) = cos 30o = $\frac{\sqrt{3}}{2}$
cos A cos B + sin A sin B = cos 60o cos 30o + sin 60o sin 30o
=
∴​ cos (AB) = cos A cos B + sin A sin B

(iii) tan (AB) = tan 30o = $\frac{1}{\sqrt{3}}$

∴​ tan (AB) =

#### Question 16:

If A and B are acute angles such that tan A = $\frac{1}{3}$, tan B$\frac{1}{2}$ and tan (A + B) =  , show that A + B = 45°.

Given:

13

#### Question 17:

Using the formula, tan , find the value of tan 60°, it being given that tan 30° = $\frac{1}{\sqrt{3}}$.

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ tan 60o = $\sqrt{3}$

#### Question 18:

Using the formula,  , find the value of cos 30°, it being given that cos 60° = $\frac{1}{2}$.

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ cos 30o = $\frac{\sqrt{3}}{2}$

#### Question 19:

Using the formula,  , find the value of sin 30°, it being given that cos 60° = $\frac{1}{2}$.

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ sin 30o = $\frac{1}{2}$

#### Question 20:

If and , find the values of .

Disclaimer: $\mathrm{cos}15°$ can also be calculated by taking .

#### Question 21:

If tan (x + 30°) = 1 then find the value of x.

As we know that,

#### Question 22:

If then find the acute angle θ.

#### Question 23:

If sin (A + B) = 1 and then find the values of A and B.

As we know that,

#### Question 24:

If then find the values of A and B.

As we know that,

#### Question 25:

If tan (A − B) = $\frac{1}{\sqrt{3}}$ and tan (A + B) = $\sqrt{3}$, 0° < (A + B) < 90° and A > B, then find A and B.

Here, tan (A − B) = $\frac{1}{\sqrt{3}}$
⇒ tan (A B) = tan 30o       [∵ tan 30o = $\frac{1}{\sqrt{3}}$]
A − B = 30o                     ...(i)

Also, tan (A + B) = $\sqrt{3}$
⇒​ tan (A + B) =  tan 60o        [∵ tan 60o = $\sqrt{3}$]
A + B = 60o                           ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 15o

#### Question 26:

If cosec (A + B) = 1 and cosec(A – B) = 2, 0° < (A + B) ≤ 90° and A > B then find the values of :
(i) sin A cos B + cos A sin B
(ii)

As we know that,

#### Question 31:

Find the value of x for which
(i) x tan 45° cot 60° = sin 30° cosec 60°
(ii)

(i) As we know that,

(ii) As we know that,

(sec260°– 1) = ?
(a) 0
(b) 2
(c) 3
(d) 4

As we know that,

#### Question 2:

(sin230° – sec260° + 4cot245°) = ?
(a) 4
(b) 2
(c) 1
(d) $\frac{1}{4}$

As we know that,

#### Question 3:

(3cos260° + 2cot230° – 5sin245°) = ?
(a) 1

(b) 4

(c) $\frac{17}{4}$

(d) $\frac{13}{6}$

As we know that,

#### Question 4:

(a) $\frac{81}{8}$

(b) $\frac{83}{8}$

(c) $\frac{73}{8}$

(d) $\frac{75}{8}$

As we know that,

#### Question 5:

(cos 0° + sin 30° + sin 45°) (sin 90° + cos 60° – cos 45°) = ?

(a) $\frac{7}{4}$

(b) $\frac{5}{6}$

(c) $\frac{3}{5}$

(d) $\frac{5}{8}$

As we know that,

#### Question 6:

If tan2 45° – cos230° = x sin 45° cos 45° then x = ?

(a) 2

(b) –2

(c) $\frac{1}{2}$

(d) $-\frac{1}{2}$

As we know that,

#### Question 7:

If $\sqrt{2}$ sin (60° – α) = 1 then α = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

As we know that,

#### Question 8:

If tan x = 3 cot x then x = ?
(a) 60°
(b) 45°
(c) 30°
(d) 15°

Given: tan= 3cotx

If
(a) 15°
(b) 30°
(c) 45°
(d) 60°

As we know that,

If
(a) 30°
(b) 45°
(c) 60°
(d) 90°

As we know that,

If
(a) 10°
(b) 15°
(c) 20°
(d) 30°

As we know that,

#### Question 12:

If x tan 45° cos 60° = sin 60°cot 60° then x = ?
(a) 1

(b) $\frac{1}{2}$

(c) $\frac{1}{\sqrt{2}}$

(d) $\sqrt{3}$

As we know that,

#### Question 13:

If tan (3x + 30°) = 1 then x = ?
(a) 20
(b) 15°
(c) 10°
(d) 5°

As we know that,

#### Question 14:

If sin θ = cos θ, 0 ≤ θ ≤ 90° then θ = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

#### Question 15:

If then values of A and B are
(a) (60°, 30°)
(b) (60°, 15°)
(c) (45°, 15°)
(d) (60°, 25°)