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#### Question 1:

sin4θ – cos4θ = 1 – 2cos2θ

#### Question 2:

We know
${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$
Squaring on both sides, we get

$\therefore \frac{{\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta }{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }=\frac{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }=1$

#### Question 3:

$1+\frac{{\mathrm{cot}}^{2}\mathrm{\theta }}{1+\mathrm{cosec\theta }}=\mathrm{cosec\theta }$

Ans

#### Question 4:

(sec A – cos A) (cot A + tan A) = sec A tan A

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

#### Question 6:

sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A

Ans

#### Question 7:

sec4θ – tan4θ = 1 + 2 tan2θ

#### Question 8:

$\frac{\mathrm{sin\theta }}{\left(\mathrm{cot\theta }+\mathrm{cosec\theta }\right)}-\frac{\mathrm{sin\theta }}{\left(\mathrm{cot\theta }-\mathrm{cosec\theta }\right)}=2$

#### Question 9:

$\frac{\mathrm{cosec}A-\mathrm{sin}A}{\mathrm{cosec}A+\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}=\frac{\frac{1}{\mathrm{sin}A}-\mathrm{sin}A}{\frac{1}{\mathrm{sin}A}+\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}=\frac{\frac{1-{\mathrm{sin}}^{2}A}{\mathrm{sin}A}}{\frac{1+{\mathrm{sin}}^{2}A}{\mathrm{sin}A}}\phantom{\rule{0ex}{0ex}}=\frac{1-{\mathrm{sin}}^{2}A}{1+{\mathrm{sin}}^{2}A}$
$=\frac{\frac{1-{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}}{\frac{1+{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}}$           (Dividing numerator and denominator by cos2A)

#### Question 10:

sin2θ tanθ + cos2θ cot θ + 2sin θ cos θ = tan θ + cot θ

${\mathrm{sin}}^{2}\theta \mathrm{tan}\theta +{\mathrm{cos}}^{2}\theta \mathrm{cot}\theta +2\mathrm{sin}\theta \mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{3}\theta }{\mathrm{cos}\theta }+\frac{{\mathrm{cos}}^{3}\theta }{\mathrm{sin}\theta }+2\mathrm{sin}\theta \mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta +2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }{\mathrm{sin}\theta \mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{{\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)}^{2}}{\mathrm{sin}\theta \mathrm{cos}\theta }$

$=\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\theta +\mathrm{cot}\theta$

#### Question 11:

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

#### Question 12:

(1 + cot A + tan A) (sin A – cos A) = sin A tan A – cot A cos A

$\left(1+\mathrm{cot}A+\mathrm{tan}A\right)\left(\mathrm{sin}A-\mathrm{cos}A\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A+\mathrm{cot}A\mathrm{sin}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\mathrm{tan}A\mathrm{cos}A\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A+\frac{\mathrm{cos}A}{\mathrm{sin}A}×\mathrm{sin}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\frac{\mathrm{sin}A}{\mathrm{cos}A}×\mathrm{cos}A$
$=\mathrm{sin}A+\mathrm{cos}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\mathrm{sin}A\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A\mathrm{tan}A-\mathrm{cot}A\mathrm{cos}A$

#### Question 13:

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

Ans

#### Question 14:

${\mathrm{sec}}^{2}\mathrm{\theta }-\frac{{\mathrm{sin}}^{2}\mathrm{\theta }-2{\mathrm{sin}}^{4}\mathrm{\theta }}{2{\mathrm{cos}}^{4}\mathrm{\theta }-{\mathrm{cos}}^{2}\mathrm{\theta }}=1$

#### Question 15:

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

Ans

#### Question 21:

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

ANS

#### Question 22:

$\frac{{\mathrm{cos}}^{3}\mathrm{\theta }+{\mathrm{sin}}^{3}\mathrm{\theta }}{\mathrm{cos\theta }+\mathrm{sin\theta }}+\frac{{\mathrm{cos}}^{3}\mathrm{\theta }-{\mathrm{sin}}^{3}\mathrm{\theta }}{\mathrm{cos\theta }-\mathrm{sin\theta }}=2$

Hence, LHS= RHS

#### Question 23:

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

Hence, L.H.S. = R.H.S.

ANS

ANS

#### Question 26:

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

Ans

#### Question 27:

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

Ans

Hence, LHS = RHS

#### Question 29:

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

Hence, LHS = RHS

#### Question 30:

Prove that $\frac{{\mathrm{cot}}^{2}\theta \left(\mathrm{sec}\theta -1\right)}{\left(1+\mathrm{sin}\theta \right)}+\frac{{\mathrm{sec}}^{2}\theta \left(\mathrm{sin}\theta -1\right)}{\left(1+\mathrm{sec}\theta \right)}=0$

#### Question 31:

Show that none of the following is an identity:
(i) cos2θ + cos θ = 1
(ii) sin2θ + sin θ = 2
(iii) tan2θ + sin θ = cos2θ

#### Question 1:

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

#### Question 2:

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

#### Question 3:

prove that $\left(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}\right)=2.$

#### Question 4:

If x = a cos3θ and y = b sin3θ, prove that ${\left(\frac{x}{a}\right)}^{2/3}+{\left(\frac{y}{b}\right)}^{2/3}=1.$

#### Question 5:

If x = b sec3θ and y = a tan3θ, prove that ${\left(\frac{x}{b}\right)}^{2/3}-{\left(\frac{y}{a}\right)}^{2/3}=1$.

Also,

We know

#### Question 6:

If (tan θ + sin θ) = m and (tan θ − sin θ) = n, prove that (m2 n2)2 = 16mn.

#### Question 7:

If (cot θ + tan θ) = m and (sec θ − cos θ) = n, prove that (m2n)2/3 − (mn2)2/3 = 1.

#### Question 8:

If (cosec θ − sin θ) = a3 and (sec θ − cos θ) = b3, prove that ${a}^{2}{b}^{2}\left({a}^{2}+{b}^{2}\right)=1.$

#### Question 9:

If x = (sec A + sin A) and y = (sec A – sin A), prove that ${\left(\frac{2}{x+y}\right)}^{2}+{\left(\frac{x-y}{2}\right)}^{2}=1$.

Given:

x = secA + sinA       .....(1)

y = secA – sinA       .....(2)

Adding (1) and (2), we get

$x+y=\mathrm{sec}A+\mathrm{sin}A+\mathrm{sec}A-\mathrm{sin}A\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sec}A=x+y\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}A=\frac{x+y}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sec}A}=\frac{2}{x+y}$

Subtracting (2) from (1), we get

We know

#### Question 10:

If , then show that $\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}=\frac{2}{\sqrt{1-{\mathrm{tan}}^{2}\theta }}$.

$\mathrm{LHS}=\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{m}}{\sqrt{n}}+\frac{\sqrt{n}}{\sqrt{m}}\phantom{\rule{0ex}{0ex}}=\frac{m+n}{\sqrt{mn}}\phantom{\rule{0ex}{0ex}}=\frac{\left(\mathrm{cos}\theta -\mathrm{sin}\theta \right)+\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)}{\sqrt{\left(\mathrm{cos}\theta -\mathrm{sin}\theta \right)\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)}}$
$=\frac{2\mathrm{cos}\theta }{\sqrt{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{2\mathrm{cos}\theta }{\mathrm{cos}\theta }\right)}{\left(\frac{\sqrt{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }}{\mathrm{cos}\theta }\right)}\phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{\frac{{\mathrm{cos}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }-\frac{{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }}}\phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{1-{\mathrm{tan}}^{2}\theta }}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Question 11:

If (cos θ – sin θ) = $\sqrt{2}$ sin θ then prove that (cos θ + sin θ) = $\sqrt{2}$ cos θ.

$\mathrm{cos}\theta -\mathrm{sin}\theta =\sqrt{2}\mathrm{sin}\theta$
Squaring on both sides, we get

#### Question 12:

If (sec θ – tan θ) = $\sqrt{2}$ tan θ then prove that (sec θ + tan θ) = $\sqrt{2}$ sec θ.

$\mathrm{sec}\theta -\mathrm{tan}\theta =\sqrt{2}\mathrm{tan}\theta$
Squaring on both sides, we get

#### Question 13:

If (sec θ + tan θ ) = p then show that (sec θ – tan θ) = $\frac{1}{p}$.
Hence, show that cos θ = .

Given:
We know

Adding (1) and (2), we get
$\mathrm{sec}\theta +\mathrm{tan}\theta +\mathrm{sec}\theta -\mathrm{tan}\theta =p+\frac{1}{p}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sec}\theta =\frac{{p}^{2}+1}{p}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}\theta =\frac{{p}^{2}+1}{2p}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sec}\theta }=\frac{2p}{{p}^{2}+1}$

Subtracting (2) from (1), we get

Now,

#### Question 14:

If (cosec A + cot A) = m then prove that .

Given:
We know

Adding (1) and (2), we get
$\mathrm{cosec}A+\mathrm{cot}A+\mathrm{cosec}A-\mathrm{cot}A=m+\frac{1}{m}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cosec}A=\frac{{m}^{2}+1}{m}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cosec}A=\frac{{m}^{2}+1}{2m}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{cosec}A}=\frac{2m}{{m}^{2}+1}$

Subtracting (2) from (1), we get

Now,

#### Question 15:

If (sec A – tan A) = x then prove that $\frac{1+{x}^{2}}{1-{x}^{2}}$= cosec A.

Given:
We know

Adding (1) and (2), we get

Subtracting (1) from (2), we get

Dividing (3) by (4), we get
$\frac{\frac{1+{x}^{2}}{2x}}{\frac{1-{x}^{2}}{2x}}=\frac{\mathrm{sec}A}{\mathrm{tan}A}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\frac{\frac{1}{\mathrm{cos}A}}{\frac{\mathrm{sin}A}{\mathrm{cos}A}}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\frac{1}{\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\mathrm{cosec}A$

#### Question 1:

If sin A + sin2 A = 1, then (cos2 A + cos4 A) = ?
(a) $\frac{1}{2}$
(b) 1
(c) 2
(d) 3

(b) 1

#### Question 2:

$\frac{1+{\mathrm{tan}}^{2}A}{1+{\mathrm{cot}}^{2}A}=?$
(a) –1
(b) sec2A
(c) tan2A
(d) cot2A

Hence, the correct answer is option (c).

#### Question 3:

(sec A + tan A) (1 − sin A) = ?
(a) sin A
(b) cos A
(c) sec A
(d) cosec A

(b) cos A

#### Question 4:

(1 + tan θ + sec θ) (1 + cot θ – cosec θ) = ?
(a) –1
(b) 0
(c) 1
(d) 2

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

Hence, the correct answer is option (d).

#### Question 5:

If sin θ – cos θ = 0 then the value of (sin4θ + cos4θ) is
(a) $\frac{1}{4}$

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

(c) $\frac{3}{4}$

(d) 1

$\mathrm{sin}\theta -\mathrm{cos}\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\theta =\mathrm{tan}45°$
$⇒\theta =45°\phantom{\rule{0ex}{0ex}}\therefore {\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta \phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{4}45°+{\mathrm{cos}}^{4}45°\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{\sqrt{2}}\right)}^{4}+{\left(\frac{1}{\sqrt{2}}\right)}^{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}+\frac{1}{4}$
$=\frac{2}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Hence, the correct answer is option (b).

#### Question 6:

If cos 9α = sin α and 9α < 90° then the value of tan 5α is

(a) $\frac{1}{\sqrt{3}}$

(b) $\sqrt{3}$

(c) 1

(d) 0

$\mathrm{cos}9\alpha =\mathrm{sin}\alpha \phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}9\alpha =\mathrm{cos}\left(90°-\alpha \right)\phantom{\rule{0ex}{0ex}}⇒9\alpha =90°-\alpha \phantom{\rule{0ex}{0ex}}⇒10\alpha =90°$
$⇒5\alpha =\frac{90°}{2}=45°\phantom{\rule{0ex}{0ex}}\therefore \mathrm{tan}5\alpha =\mathrm{tan}45°=1$

Thus, the value of tan5α is 1.

Hence, the correct answer is option (c).

#### Question 7:

The value of $\frac{{\mathrm{tan}}^{2}\mathrm{\theta }-{\mathrm{sec}}^{2}\mathrm{\theta }}{{\mathrm{cot}}^{2}\mathrm{\theta }-{\mathrm{cosec}}^{2}\mathrm{\theta }}$ is
(a) –1

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

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

(d) 1

Hence, the correct answer is option (d).

#### Question 8:

(a) (sec A + tan A)
(b) (sec A − tan A)
(c) sec A tan A
(d) None to these

(b) (sec A − tan A)

#### Question 9:

(a) cosec A – cot A
(b) cosec A + cot A
(c) cosec A cot A
(d) none of these

$\sqrt{\frac{1+\mathrm{cos}\left(A\right)}{1-\mathrm{cos}\left(A\right)}}=\sqrt{\frac{1+\mathrm{cos}\left(A\right)}{1-\mathrm{cos}\left(A\right)}×\frac{1+\mathrm{cos}\left(A\right)}{1+\mathrm{cos}\left(A\right)}}=\frac{1+\mathrm{cos}\left(A\right)}{\sqrt{1-{\mathrm{cos}}^{2}\left(A\right)}}\phantom{\rule{0ex}{0ex}}=\frac{1+\mathrm{cos}\left(A\right)}{\sqrt{{\mathrm{sin}}^{2}\left(A\right)}}=\frac{1+\mathrm{cos}\left(A\right)}{\mathrm{sin}\left(A\right)}=\frac{1}{\mathrm{sin}\left(A\right)}+\frac{\mathrm{cos}\left(A\right)}{\mathrm{sin}\left(A\right)}\phantom{\rule{0ex}{0ex}}=\mathrm{cosec}\left(A\right)+\mathrm{cot}\left(A\right).$
Hence, the correct answer is option B.

#### Question 10:

(cosec θ − cot θ)2 = ?
(a) $\frac{1+\mathrm{cos\theta }}{1-\mathrm{cos\theta }}$
(b) $\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}$
(c)  $\frac{1+\mathrm{sin\theta }}{1-\mathrm{sin\theta }}$
(d) $\frac{1-\mathrm{sin\theta }}{1+\mathrm{sin\theta }}$

(b) $\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}$

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

#### Question 11:

If (tan θ + cot θ) = 5, then (tan2 θ + cot2 θ) = ?

(a) 23
(b) 24
(c) 25
(d) 27

(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)2 = 52
⇒ tan2 θ + cot2 θ + 2 tan θ cot θ = 25
⇒ tan2 θ + cot2 θ + 2 = 25       [∵ tan θ = ]
⇒ tan2 θ + cot2 θ = 25 − 2 = 23

#### Question 12:

If tan θ = $\frac{a}{b},$ then $\frac{\left(\mathrm{cos\theta }+\mathrm{sin\theta }\right)}{\left(\mathrm{cos\theta }-\mathrm{sin\theta }\right)}=?$
(a) $\frac{a+b}{a-b}$
(b) $\frac{a-b}{a+b}$
(c) $\frac{b+a}{b-a}$
(d) $\frac{b-a}{b+a}$

(c) $\frac{b+a}{b-a}$

#### Question 13:

If tan θ = $\frac{1}{\sqrt{7}}$, then $\frac{\left(\mathrm{cos}e{c}^{2}\mathrm{\theta }-se{c}^{2}\mathrm{\theta }\right)}{\left(\mathrm{cos}e{c}^{2}\mathrm{\theta }+se{c}^{2}\mathrm{\theta }\right)}=?$ = ?
(a) $\frac{-2}{3}$
(b) $\frac{-3}{4}$
(c) $\frac{2}{3}$
(d) $\frac{3}{4}$

(d) $\frac{3}{4}$

=

#### Question 14:

If tan $\mathrm{\theta }=\frac{8}{15}$ then cosec θ = ?
(a) $\frac{17}{8}$

(b) $\frac{8}{17}$

(c) $\frac{15}{17}$

(d) $\frac{17}{15}$

Now,

Hence, the correct answer is option (a).

#### Question 15:

If
(a) $\frac{2}{3}$

(b) $\frac{1}{3}$

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

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

Hence, the correct answer is option (c).

#### Question 16:

If in ΔABC, ∠C = 90°. Then, the value of cos(A + B) is

(a) 0

(b) 1

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

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

In ∆ABC,

∠A + ∠B + ∠C = 180º                (Angle sum property of triangle)

⇒ ∠A + ∠B + 90º = 180º            (∠C = 90º)

⇒ ∠A + ∠B = 180º − 90º = 90º

∴ cos(A + B) = cos90º = 0

Thus, the value of cos(A + B) is 0.

Hence, the correct answer is option (a).

#### Question 17:

If cos A + cos2 A = 1, then (sin2 A + sin4 A) = ?
(a) 1
(b) 2
(c) 3
(d) 4

(a) 1

#### Question 18:

If 3 cot θ = 4, then $\frac{\left(5\mathrm{sin\theta }+3\mathrm{cos\theta }\right)}{\left(5\mathrm{sin\theta }-3\mathrm{cos\theta }\right)}$ = ?
(a) $\frac{1}{3}$
(b) 3
(c) $\frac{1}{9}$
(d) 9

(d) 9

We have .

Dividing the numerator and denominator of the given expression by sin θ, we get:

=

= = 9              [∵ 3 cot θ = 4]

#### Question 19:

If 2x = sec A and $\frac{2}{x}$ = tan A, then $2\left({x}^{2}-\frac{1}{{x}^{2}}\right)=?$ =?
(a) $\frac{1}{2}$
(b) $\frac{1}{4}$
(c) $\frac{1}{8}$
(d) $\frac{1}{16}$

(a) $\frac{1}{2}$
Given: 2x = sec A and $\frac{2}{x}$ = tan A
Also, we can deduce that x = and .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
2$\left({x}^{2}-\frac{1}{{x}^{2}}\right)$ = 2
= 2
=
= $\frac{1}{2}$                  [By using the identity: ]

#### Question 20:

If 3x = cosec θ and $\frac{3}{x}$ = cot θ, than $3\left({x}^{2}-\frac{1}{{x}^{2}}\right)=?$
(a) $\frac{1}{27}$
(b) $\frac{1}{81}$
(c) $\frac{1}{3}$
(d) $\frac{1}{9}$

(c) $\frac{1}{3}$
Given: 3x = cosec θ and $\frac{3}{x}$ = cot θ
Also, we can deduce that x = and .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
3$\left({x}^{2}-\frac{1}{{x}^{2}}\right)$ = 3
= 3
=
= $\frac{1}{3}$       [By using the identity: ]

#### Question 21:

If tan θ = $\sqrt{3}$, then sec θ = ?
(a) $\frac{2}{\sqrt{3}}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) 2

(d) 2

Let us first draw a right $∆$ABC right angled at B and $\angle \mathrm{A}=\theta$.
Given: tan θ = $\sqrt{3}$
But tan θ = $\frac{BC}{AB}$
So, $\frac{BC}{AB}$ = $\frac{\sqrt{3}}{1}$
Thus, BC = $\sqrt{3}$k and AB = k

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = ($\sqrt{3}$ k)2 + (k)2
⇒ AC2= 4k2
⇒ AC = 2k
∴ sec θ =

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