Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 8 Trigonometric Identities are provided here with simple step-by-step explanations. These solutions for Trigonometric Identities are extremely popular among Class 10 students for Math Trigonometric Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Question 1:

(i) (1 − cos2θ) cosec2θ = 1
(ii) (1 + cot2θ) sin2θ = 1

#### Question 2:

(i) (sec2θ − 1) cot2θ = 1
(ii) (sec2θ − 1) (cosec2θ − 1) = 1
(iii) (1− cos2θ) sec2θ = tan2θ

#### Question 3:

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

#### Question 4:

(i) (1 + cos θ) (1 − cos θ) (1 + cot2θ) = 1
(ii) cosec θ (1 + cos θ) (cosec θ − cot θ) = 1

#### Question 5:

Prove each of the following identities:

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

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

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

#### Question 6:

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

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

#### Question 7:

(i) sec θ (1 − sin θ) (sec θ + tan θ) = 1
(ii) sin θ(1 + tan θ) + cos θ(1 + cot θ) = (sec θ + cosec θ)

#### Question 8:

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

#### Question 9:

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

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

#### Question 10:

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

Hence, LHS = RHS

#### Question 11:

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

Hence, LHS = RHS

#### Question 13:

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

#### Question 14:

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

Hence, LHS = RHS

#### Question 15:

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

Hence, LHS = RHS

Hence, LHS = RHS

#### Question 17:

(i) ${\mathrm{sin}}^{6}\mathrm{\theta }+{\mathrm{cos}}^{6}\mathrm{\theta }=1-3{\mathrm{sin}}^{2}{\mathrm{\theta cos}}^{2}\mathrm{\theta }$
(ii) ${\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{4}\mathrm{\theta }={\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{4}\mathrm{\theta }$
(iii) ${\mathrm{cosec}}^{4}\mathrm{\theta }-{\mathrm{cosec}}^{2}\mathrm{\theta }={\mathrm{cot}}^{4}\mathrm{\theta }+{\mathrm{cot}}^{2}\mathrm{\theta }$

#### Question 18:

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

#### Question 19:

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

#### Question 20:

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

#### Question 21:

Prove each of the following identities:

#### 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:

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

#### Question 24:

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

#### Question 25:

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

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

(i)
(ii)

#### Question 27:

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

#### Question 28:

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

Hence, LHS = RHS

#### Question 29:

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

Hence, LHS = RHS

#### Question 31:

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

Hence, LHS = RHS

#### Question 32:

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

Hence, LHS = RHS

#### Question 36:

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 37:

Prove that $\left(\mathrm{sin}\theta -2{\mathrm{sin}}^{3}\theta \right)=\left(2{\mathrm{cos}}^{3}\theta -\mathrm{cos}\theta \right)\mathrm{tan}\theta$

$\mathrm{RHS}=\left(2{\mathrm{cos}}^{3}\theta -\mathrm{cos}\theta \right)\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}=\left(2{\mathrm{cos}}^{2}\theta -1\right)\mathrm{cos}\theta ×\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\left[2\left(1-{\mathrm{sin}}^{2}\theta \right)-1\right]\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}=\left(2-2{\mathrm{sin}}^{2}\theta -1\right)\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}=\left(1-2{\mathrm{sin}}^{2}\theta \right)\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}=\left(\mathrm{sin}\theta -2{\mathrm{sin}}^{3}\theta \right)\phantom{\rule{0ex}{0ex}}=\mathrm{LHS}$

#### 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 (sec θ + tan θ) = m and (sec θ − tan θ) = n, show that mn = 1.

#### Question 5:

If (cosec θ + cot θ) = m and (cosec θ − cot θ) = n, show that mn = 1.

#### Question 6:

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 7:

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

#### Question 8:

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

#### Question 9:

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 10:

If (2 sin θ + 3 cos θ) = 2, show that (3 sin θ − 2 cos θ) = ± 3.

#### Question 11:

If $\left(\mathrm{sin}\theta +\mathrm{cos}\theta \right)=\sqrt{2}\mathrm{cos}\theta$, show that $\mathrm{cot}\theta =\left(\sqrt{2}+1\right)$.

#### Question 12:

If cos θ + sin θ = $\sqrt{2}$ sin θ, show that sin θ − cos θ = $\sqrt{2}$ cos θ.

#### Question 13:

If $\mathrm{sec}\theta +\mathrm{tan}\theta =p$, prove that

#### Question 14:

If tan A = n tan B and sin A = m sin B, prove that cos2A = $\frac{\left({m}^{2}-1\right)}{{n}^{2}-1}.$

#### Question 15:

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 1:

Write the value of $\left(1-{\mathrm{sin}}^{2}\theta \right){\mathrm{sec}}^{2}\theta$.

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

#### Question 2:

Write the value of $\left(1-{\mathrm{cos}}^{2}\theta \right){\mathrm{cosec}}^{2}\theta$.

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

#### Question 3:

Write the value of $\left(1+{\mathrm{tan}}^{2}\theta \right){\mathrm{cos}}^{2}\theta$.

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

#### Question 4:

Write the value of $\left(1+{\mathrm{cot}}^{2}\theta \right){\mathrm{sin}}^{2}\theta$.

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

#### Question 5:

Write the value of $\left({\mathrm{sin}}^{2}\theta +\frac{1}{1+{\mathrm{tan}}^{2}\theta }\right)$.

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

#### Question 6:

Write the value of $\left({\mathrm{cot}}^{2}\theta -\frac{1}{{\mathrm{sin}}^{2}\theta }\right)$.

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

#### Question 7:

Write the value of .

#### Question 8:

Write the value of ${\mathrm{cosec}}^{2}\left(90°-\theta \right)-{\mathrm{tan}}^{2}\theta$.

${\mathrm{cosec}}^{2}\left(90°-\theta \right)-{\mathrm{tan}}^{2}\theta \phantom{\rule{0ex}{0ex}}={\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta \phantom{\rule{0ex}{0ex}}=1$

#### Question 9:

Write the value of ${\mathrm{sec}}^{2}\theta \left(1+\mathrm{sin}\theta \right)\left(1-\mathrm{sin}\theta \right)$.                   [CBSE 2009]

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

#### Question 10:

Write the value of $\mathrm{cos}{\mathrm{ec}}^{2}\theta \left(1+\mathrm{cos}\theta \right)\left(1-\mathrm{cos}\theta \right)$.

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

Disclaimer: The question given in the textbook is incorrect. There should be $\mathrm{cos}\theta$ instead $\mathrm{sin}\theta$. The solution provided here is of the same.

#### Question 11:

Write the value of .

#### Question 12:

Write the value of $\left(1+{\mathrm{tan}}^{2}\theta \right)\left(1+\mathrm{sin}\theta \right)\left(1-\mathrm{sin}\theta \right)$.                [CBSE 2008]

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

#### Question 13:

Write the value of $3{\mathrm{cot}}^{2}\theta -3{\mathrm{cosec}}^{2}\theta$.

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

#### Question 14:

Write the value of $4{\mathrm{tan}}^{2}\theta -\frac{4}{{\mathrm{cos}}^{2}\theta }$.

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

#### Question 15:

Write the value of $\frac{{\mathrm{tan}}^{2}\theta -{\mathrm{sec}}^{2}\theta }{{\mathrm{cot}}^{2}\theta -{\mathrm{cosec}}^{2}\theta }$.

$\frac{{\mathrm{tan}}^{2}\theta -{\mathrm{sec}}^{2}\theta }{{\mathrm{cot}}^{2}\theta -{\mathrm{cosec}}^{2}\theta }\phantom{\rule{0ex}{0ex}}=\frac{-1}{-1}\phantom{\rule{0ex}{0ex}}=1$

Now,

#### Question 17:

$4+4{\mathrm{tan}}^{2}\theta \phantom{\rule{0ex}{0ex}}=4\left(1+{\mathrm{tan}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}=4{\mathrm{sec}}^{2}\theta \phantom{\rule{0ex}{0ex}}=\frac{4}{{\mathrm{cos}}^{2}\theta }\phantom{\rule{0ex}{0ex}}=\frac{4}{{\left(\frac{2}{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4}{\left(\frac{4}{9}\right)}\phantom{\rule{0ex}{0ex}}=\frac{4×9}{4}\phantom{\rule{0ex}{0ex}}=9$

#### Question 19:

$\frac{\mathrm{sec}\theta -1}{\mathrm{sec}\theta +1}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{1}{\mathrm{cos}\theta }-\frac{1}{1}\right)}{\left(\frac{1}{\mathrm{cos}\theta }+\frac{1}{1}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{1-\mathrm{cos}\theta }{\mathrm{cos}\theta }\right)}{\left(\frac{1+\mathrm{cos}\theta }{\mathrm{cos}\theta }\right)}\phantom{\rule{0ex}{0ex}}=\frac{1-\mathrm{cos}\theta }{1+\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{1}{1}-\frac{2}{3}\right)}{\left(\frac{1}{1}+\frac{2}{3}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{1}{3}\right)}{\left(\frac{5}{3}\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{5}$

#### Question 23:

$\frac{\left({\mathrm{cosec}}^{2}\theta -{\mathrm{sec}}^{2}\theta \right)}{\left({\mathrm{cosec}}^{2}\theta +{\mathrm{sec}}^{2}\theta \right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+{\mathrm{cot}}^{2}\theta \right)-\left(1+{\mathrm{tan}}^{2}\theta \right)}{\left(1+{\mathrm{cot}}^{2}\theta \right)+\left(1+{\mathrm{tan}}^{2}\theta \right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\frac{1}{{\mathrm{tan}}^{2}\theta }\right)-\left(1+{\mathrm{tan}}^{2}\theta \right)}{\left(1+\frac{1}{{\mathrm{tan}}^{2}\theta }\right)+\left(1+{\mathrm{tan}}^{2}\theta \right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\frac{1}{{\mathrm{tan}}^{2}\theta }-1-{\mathrm{tan}}^{2}\theta \right)}{\left(1+\frac{1}{{\mathrm{tan}}^{2}\theta }+1+{\mathrm{tan}}^{2}\theta \right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{1}{{\mathrm{tan}}^{2}\theta }-{\mathrm{tan}}^{2}\theta \right)}{\left(\frac{1}{{\mathrm{tan}}^{2}\theta }+{\mathrm{tan}}^{2}\theta +2\right)}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\frac{\sqrt{5}}{1}\right)}^{2}-{\left(\frac{1}{\sqrt{5}}\right)}^{2}}{{\left(\frac{\sqrt{5}}{1}\right)}^{2}+{\left(\frac{1}{\sqrt{5}}\right)}^{2}+2}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{5}{1}-\frac{1}{5}\right)}{\left(\frac{5}{1}+\frac{1}{5}+\frac{2}{1}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{24}{5}\right)}{\left(\frac{36}{5}\right)}\phantom{\rule{0ex}{0ex}}=\frac{24}{36}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}$

#### Question 30:

$\left(\mathrm{sin}A+\mathrm{cos}A\right)\mathrm{sec}A\phantom{\rule{0ex}{0ex}}=\left(\mathrm{sin}A+\mathrm{cos}A\right)\frac{1}{\mathrm{cos}A}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}A}{\mathrm{cos}A}+\frac{\mathrm{cos}A}{\mathrm{cos}A}\phantom{\rule{0ex}{0ex}}=\mathrm{tan}A+1\phantom{\rule{0ex}{0ex}}=\frac{5}{12}+\frac{1}{1}\phantom{\rule{0ex}{0ex}}=\frac{5+12}{12}\phantom{\rule{0ex}{0ex}}=\frac{17}{12}$

#### Question 34:

$\left({b}^{2}{x}^{2}+{a}^{2}{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={b}^{2}{\left(a\mathrm{sin}\theta \right)}^{2}+{a}^{2}{\left(b\mathrm{cos}\theta \right)}^{2}\phantom{\rule{0ex}{0ex}}={b}^{2}{a}^{2}{\mathrm{sin}}^{2}\theta +{a}^{2}{b}^{2}{\mathrm{cos}}^{2}\theta \phantom{\rule{0ex}{0ex}}={a}^{2}{b}^{2}\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}={a}^{2}{b}^{2}\left(1\right)\phantom{\rule{0ex}{0ex}}={a}^{2}{b}^{2}$

#### Question 35:

$5\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{25}{5}\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\left(25{x}^{2}-\frac{25}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\left[{\left(5x\right)}^{2}-{\left(\frac{5}{x}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\left[{\left(\mathrm{sec}\theta \right)}^{2}-{\left(\mathrm{tan}\theta \right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\left({\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\left(1\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{5}$

#### Question 36:

$2\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{4}{2}\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(4{x}^{2}-\frac{4}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[{\left(2x\right)}^{2}-{\left(\frac{2}{x}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[{\left(\mathrm{cosec}\theta \right)}^{2}-{\left(\mathrm{sec}\theta \right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left({\mathrm{cosec}}^{2}\theta -{\mathrm{sec}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(1\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

#### Question 39:

$\mathrm{cot}\theta =\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{1-{\mathrm{sin}}^{2}\theta }}{\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{1-{x}^{2}}}{2}$

#### Question 2:

$\frac{\mathrm{tan}35°}{\mathrm{cot}55°}+\frac{\mathrm{cot}78°}{\mathrm{tan}12°}=?$
(a) 0
(b) 1
(c) 2
(d) None of these

#### Question 3:

tan 10° tan 15° tan 75° tan 80° = ?
(a) $\sqrt{3}$
(b) $\frac{1}{\sqrt{3}}$
(c) −1
(d) 1

#### Question 4:

tan 5° tan 25° tan 30° tan 65° tan 85° = ?
(a) $\sqrt{3}$
(b) $\frac{1}{\sqrt{3}}$
(c) 1
(d) none of these

#### Question 5:

cos 1° cos 2° cos 3° ... cos 180° = ?
(a) −1
(b) 1
(c) 0
(d) $\frac{1}{2}$

#### Question 6:

$\frac{2{\mathrm{sin}}^{2}63°+1+2{\mathrm{sin}}^{2}27°}{3{\mathrm{cos}}^{2}17°-2+3{\mathrm{cos}}^{2}73°}=?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 2
(d) 3

#### Question 7:

(sin 43°cos 47° + cos 43°sin 47°) = ?
(a) sin 4°
(b) cos 4°
(c) 1
(d) 0

#### Question 9:

If sin 3A = cos (A − 10°), where 3A is an acute angle, then ∠A = ?
(a) 35°
(b) 25°
(c) 20°
(d) 45°

#### Question 12:

If cos(α + β) = 0, then sin(α − β) = ?
(a) sin α
(b) cos β
(c) sin 2α
(d) cos 2β

(d) cos 2β

#### Question 13:

sin (45° + θ) − cos (45° − θ) = ?
(a) 2 sin θ
(b) 2 cos θ
(c) 0
(d) 1

#### Question 14:

(sin 79° cos 11° + cos 79° sin 11°) = ?
(a) $\frac{1}{\sqrt{2}}$
(b) $\frac{1}{2}$
(c) 0
(d) 1

#### Question 15:

(cosec257° − tan233°) = ?
(a) 0
(b) 1
(c) 2
(d) None of these

#### Question 16:

$\frac{2{\mathrm{tan}}^{2}30°{\mathrm{sec}}^{2}52°{\mathrm{sin}}^{2}38°}{\left({\mathrm{cosec}}^{2}70°-{\mathrm{tan}}^{2}20°\right)}=?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 2
(d) $\frac{1}{2}$

(b) $\frac{2}{3}$

#### Question 17:

$\left\{\frac{\left({\mathrm{sin}}^{2}22°+{\mathrm{sin}}^{2}68°\right)}{\left({\mathrm{cos}}^{2}22°+{\mathrm{cos}}^{2}68°\right)}+{\mathrm{sin}}^{2}63°+\mathrm{cos}63°\mathrm{sin}27\right\}=?$
(a) 0
(b) 1
(c) 2
(d) 3

#### Question 18:

$\frac{\mathrm{cot}\left(90°-\mathrm{\theta }\right)·\mathrm{sin}\left(90°-\mathrm{\theta }\right)}{\mathrm{sin\theta }}+\frac{\mathrm{cot}40°}{\mathrm{tan}50°}-\left({\mathrm{cos}}^{2}20°+{\mathrm{cos}}^{2}70°\right)=?$
(a) 3
(b) 2
(c) 1
(d) 0

#### Question 19:

(a) $\sqrt{3}$
(b) $\frac{1}{3}$
(c) $\frac{1}{\sqrt{3}}$
(d) $\frac{2}{\sqrt{3}}$

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

#### Question 20:

If 2 sin 2θ = $\sqrt{3}$, then θ = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

(a) 30o

#### Question 21:

If 2cos 3θ = 1, then θ = ?
(a) 10°
(b) 15°
(c) 20°
(d) 30°

(c) 20o

#### Question 22:

If $\sqrt{3}$ tan 2θ − 3 = 0, then θ = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

(b) 30o

#### Question 23:

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

(b) 60o

#### Question 24:

If x tan 45° cos 60° = sin 60° cot 60°, then x = ?
(a) 1
(b) $\frac{1}{2}$
(c) $\frac{1}{\sqrt{2}}$
(d) $\sqrt{3}$

(a) 1

#### Question 25:

If (tan245° − cos230°) = x sin 45° cos 45°, then x = ?
(a) 2
(b) −2
(c) $\frac{1}{2}$
(d) $\frac{-1}{2}$

(c) $\frac{1}{2}$

#### Question 26:

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

(b) 3

sec260o − 1 = (2)2 − 1 = 4 − 1 = 3

#### Question 27:

(cos 0° + sin 30° + sin 45°) (sin 90° + cos 60° − cos 45°) = ?
(a) $\frac{5}{6}$
(b) $\frac{5}{8}$
(c) $\frac{3}{5}$
(d) $\frac{7}{4}$

(d) $\frac{7}{4}$

#### Question 28:

(sin230° + 4cot245° − sec260°) = ?
(a) 0
(b) $\frac{1}{4}$
(c) 4
(d) 1

(b) $\frac{1}{4}$

#### Question 29:

(3cos260° + 2cot230° − 5sin245°) = ?
(a) $\frac{13}{6}$
(b) $\frac{17}{4}$
(c) 1
(d) 4

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

#### Question 30:

${\mathrm{cos}}^{2}30°{\mathrm{cos}}^{2}45°+4{\mathrm{sec}}^{2}60°+\frac{1}{2}{\mathrm{cos}}^{2}90°-2{\mathrm{tan}}^{2}60°=?$
(a) $\frac{73}{8}$
(b) $\frac{75}{8}$
(c) $\frac{81}{8}$
(d) $\frac{83}{8}$

(d) $\frac{83}{8}$

#### Question 31:

If cosec θ = $\sqrt{10}$, then sec θ = ?
(a) $\frac{3}{\sqrt{10}}$
(b) $\frac{\sqrt{10}}{3}$
(c) $\frac{1}{\sqrt{10}}$
(d) $\frac{2}{\sqrt{10}}$

(b) $\frac{\sqrt{10}}{3}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
Given: cosec $\theta$$\sqrt{10}$, but sin $\theta$ = $\frac{1}{\sqrt{10}}$
Also, sin $\theta$ = $\frac{\mathrm{Perpendicular}}{\mathrm{Hypotenuse}}$ = $\frac{BC}{AC}$
So, $\frac{BC}{AC}$ = $\frac{1}{\sqrt{10}}$
Thus, BC = k and AC = $\sqrt{10}$ k Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
AB2 = AC2 $-$ BC2
AB2 = ($\sqrt{10}$ k)2 $-$ (k)2
AB2 = 9k2
AB = 3k
∴ sec $\theta$ = $\frac{AC}{AB}$ = $\frac{\sqrt{10}k}{3k}=\frac{\sqrt{10}}{3}$

#### Question 32:

If tan θ = $\frac{8}{15}$, then cosec θ = ?
(a) $\frac{17}{8}$
(b) $\frac{8}{17}$
(c) $\frac{17}{15}$
(d) $\frac{15}{17}$

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

Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
Given: tan $\theta$ = $\frac{8}{15}$, but tan $\theta$ = $\frac{BC}{AB}$
So, $\frac{BC}{AB}$ = $\frac{8}{15}$
Thus, BC = 8k and AB = 15k Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AC2 = (15k)2 + (8k)2
⇒ AC2 = 289k2
⇒ AC = 17k
∴ cosec $\theta$ = $\frac{AC}{BC}=\frac{17k}{8k}=\frac{17}{8}$

#### Question 33:

If sin θ $\frac{a}{b}$, then cos θ = ?
(a) $\frac{b}{\sqrt{{b}^{2}-{a}^{2}}}$
(b) $\frac{\sqrt{{b}^{2}-{a}^{2}}}{b}$
(c) $\frac{a}{\sqrt{{b}^{2}-{a}^{2}}}$
(d) $\frac{b}{a}$

(b) $\frac{\sqrt{{b}^{2}-{a}^{2}}}{b}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
Given: sin θ = $\frac{a}{b}$, but sin θ = $\frac{BC}{AC}$
So, $\frac{BC}{AC}$ = $\frac{a}{b}$
Thus, BC = ak and AC = bk Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 $-$ BC2
⇒ AB2 = (bk)2 $-$ (ak)2
⇒ AB2 = $\left({b}^{2}-{a}^{2}\right)$k2
⇒ AB = ($\sqrt{{b}^{2}-{a}^{2}}$)k
∴ cos θ  = $\frac{\mathrm{AB}}{\mathrm{AC}}$ = $\frac{\sqrt{{b}^{2}-{a}^{2}}k}{bk}$ = $\frac{\sqrt{{b}^{2}-{a}^{2}}}{b}$

#### Question 34:

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 =

#### Question 35:

If sec θ = $\frac{25}{7}$, then sin θ = ?
(a) $\frac{7}{24}$
(b) $\frac{24}{7}$
(c) $\frac{24}{25}$
(d) None of these

(c) $\frac{24}{25}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
Given: sec θ = $\frac{25}{7}$
But cos θ$\frac{AB}{AC}$ = $\frac{7}{25}$
Thus, AC = 25k and AB = 7k Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
BC2 = AC2 $-$AB2
BC2 = (25k)2 $-$ (7k)2
BC2= 576k2
BC = 24k

∴ sin θ =

#### Question 36:

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

(b) $\sqrt{3}$

Given: sinθ = $\frac{1}{2}$, but sinθ = $\frac{\mathrm{BC}}{\mathrm{AC}}$
So, $\frac{\mathrm{BC}}{\mathrm{AC}}$ = $\frac{1}{2}$
Thus, BC = k and AC = 2k Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
AB2 = AC2 $-$ BC2
AB2 = (2k)2 $-$ (k)2
AB2 = 3k2
AB = $\sqrt{3}$k
So, tanθ = $\frac{\mathrm{BC}}{\mathrm{AB}}$ = $\frac{k}{\sqrt{3}k}$ = $\frac{1}{\sqrt{3}}$
∴ cotθ = $\frac{1}{\mathrm{tan}\theta }$ = $\sqrt{3}$

#### Question 37:

If cosθ = $\frac{4}{5}$, then tanθ = ?
(a) $\frac{3}{4}$
(b) $\frac{4}{3}$
(c) $\frac{3}{5}$
(d) $\frac{5}{3}$

(a) $\frac{3}{4}$
Since cosθ = $\frac{4}{5}$ but cosθ = $\frac{AB}{AC}$
So, $\frac{AB}{AC}$ = $\frac{4}{5}$
Thus, AB = 4k and AC = 5k Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 $-$ AB2
⇒ BC2  = (5k)2 $-$ (4k)2
⇒ BC2 = 9k2
⇒ BC = 3k
∴ tanθ = $\frac{BC}{AB}$ = $\frac{3}{4}$

#### Question 38:

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 39:

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 40:

If tan θ = $\frac{4}{3}$, then (sinθ + cosθ) = ?
(a) $\frac{7}{3}$
(b) $\frac{7}{4}$
(c) $\frac{7}{5}$
(d) $\frac{5}{7}$

(c) $\frac{7}{5}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
tan θ =
So, AB = 3k and BC = 4k Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
AC2 = (3k)2 + (4k)2
AC2 = 25k2
AC= 5k
Thus, sin θ = $\frac{BC}{AC}$ = $\frac{4}{5}$
and cos θ =
∴ (sin θ + cos θ) = ( $\frac{4}{5}$ + $\frac{3}{5}$) = $\frac{7}{5}$

#### Question 41:

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 42:

If (cos θ + sec θ) = $\frac{5}{2}$, then (cos2 θ + sec2 θ) = ?
(a) $\frac{21}{4}$
(b) $\frac{17}{4}$
(c) $\frac{29}{4}$
(d) $\frac{33}{4}$

(b) $\frac{17}{4}$
We have (cos θ +sec θ) = $\frac{5}{2}$
Squaring both sides, we get:
(cos θ + sec θ)2 = ($\frac{5}{2}$)2
$⇒$cos2 θ + sec2 θ + 2 cos θ sec θ = $\frac{25}{4}$
$⇒$cos2 θ + sec2 θ + 2 = $\frac{25}{4}$      [∵ sec θ = ]
$⇒$cos2 θ + sec2 θ = $\frac{25}{4}$ − 2 = $\frac{17}{4}$

#### Question 43:

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 44:

If 7 tan θ = 4, then $\frac{\left(7\mathrm{sin\theta }-3\mathrm{cos\theta }\right)}{\left(7\mathrm{sin\theta }+3\mathrm{cos\theta }\right)}$ = ?
(a) $\frac{1}{7}$
(b) $\frac{5}{7}$
(c) $\frac{3}{7}$
(d) $\frac{5}{14}$

(a) $\frac{1}{7}$

7 tan θ = 4

Now, dividing the numerator and denominator of the given expression by cos θ, we get:

=

=         [∵ 7 tan θ = 4]
$\frac{1}{7}$

#### Question 45:

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 46:

If θ = $\frac{a}{b}$, then  = ?
(a) $\frac{\left({a}^{2}+{b}^{2}\right)}{\left({a}^{2}-{b}^{2}\right)}$
(b) $\frac{\left({a}^{2}-{b}^{2}\right)}{\left({a}^{2}+{b}^{2}\right)}$
(c) $\frac{{a}^{2}}{\left({a}^{2}+{b}^{2}\right)}$
(d) $\frac{{b}^{2}}{\left({a}^{2}+{b}^{2}\right)}$

(b)
We have tan θ = $\frac{a}{b}$

Now, dividing the numerator and denominator of the given expression by cos θ, we get:

#### Question 47:

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

(b) 1

#### Question 48:

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

(a) 1

#### Question 49:

(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 50:

(a) (cosec A − cot A)
(b) (cosec A + cot A)
(c) cosec A cot A
(d) None of these

(a) (cosec A − cot A)

#### Question 51:

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 52:

(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 53:

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

(b) cos A

#### Question 1:

(a) $3\frac{1}{2}$
(b) 4
(c) 6
(d) 5

(b) 4

#### Question 2:

The value of $\left({\mathrm{sin}}^{2}30°{\mathrm{cos}}^{2}45°+4{\mathrm{tan}}^{2}30°+\frac{1}{2}{\mathrm{sin}}^{2}90°+\frac{1}{8}{\mathrm{cot}}^{2}60°\right)=?$
(a) $\frac{3}{8}$
(b) $\frac{5}{8}$
(c) 6
(d) 2

(d) 2

#### Question 3:

If cos A + cos2 A = 1, then (sin2 A + sin4 A) = ?
(a) $\frac{1}{2}$
(b) 2
(c) 1
(d) 4

#### Question 4:

If sin $\mathrm{\theta }=\frac{\sqrt{3}}{2}$, then (cosec θ + cot θ) = ?
(a) $\left(2+\sqrt{3}\right)$
(b) $2\sqrt{3}$
(c) $\sqrt{2}$
(d) $\sqrt{3}$

(d) $\sqrt{3}$

#### Question 5:

If cot $A=\frac{4}{5}$, prove that $\frac{\left(\mathrm{sin}A+\mathrm{cos}A\right)}{\left(\mathrm{sin}A-\mathrm{cos}A\right)}=9.$

#### Question 6:

If 2x = sec A and $\frac{2}{x}$ = tan A, prove that $\left({x}^{2}-\frac{1}{{x}^{2}}\right)=\frac{1}{4}.$

#### Question 7:

If $\sqrt{3}$ tan θ = 3 sin θ, prove that (sin2 θ − cos2 θ) = $\frac{1}{3}.$

#### Question 8:

Prove that $\frac{\left({\mathrm{sin}}^{2}73°+{\mathrm{sin}}^{2}17°\right)}{\left({\mathrm{cos}}^{2}28°+{\mathrm{cos}}^{2}62°\right)}=1.$

$\frac{\left({\mathrm{sin}}^{2}73°+{\mathrm{sin}}^{2}17°\right)}{\left({\mathrm{cos}}^{2}28°+{\mathrm{cos}}^{2}62°\right)}=1.$

#### Question 9:

If 2 sin 2θ =$\sqrt{3}$, prove that θ = 30°.

#### Question 10:

Prove that  = (cosec A + cot A).

= (cosec A + cot A).

#### Question 11:

If cosec θ + cot θ = p, prove that cos θ = $\frac{\left({p}^{2}-1\right)}{\left({p}^{2}+1\right)}.$

#### Question 12:

Prove that (cosec A − cot A)2 =

(cosec A − cot A)2 =

#### Question 13:

If 5 cot θ = 3, find the value of $\left(\frac{5\mathrm{sin\theta }-3\mathrm{cos\theta }}{4\mathrm{sin\theta }+3\mathrm{cos\theta }}\right).$

#### Question 14:

Prove that (sin 32° cos 58° + cos 32° sin 58°) = 1.

(sin 32° cos 58° + cos 32° sin 58°) = 1

#### Question 15:

If x = a sin θ + b cos 0 and y = a cos 0 b sin θ, prove that ${x}^{2}+{y}^{2}={a}^{2}+{b}^{2}.$

#### Question 16:

Prove that $\frac{\left(1+\mathrm{sin\theta }\right)}{\left(1-\mathrm{sin\theta }\right)}$ = (sec θ + tan θ)2.

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

#### Question 17:

Prove that $\frac{1}{\left(\mathrm{sec\theta }-\mathrm{tan\theta }\right)}-\frac{1}{\mathrm{cos\theta }}=\frac{1}{\mathrm{cos\theta }}-\frac{1}{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}$.

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

Prove that

Prove that

#### Question 20:

If sec 5 A = cosec (A − 36) and 5 A is an acute angle, find the value of A.

#### Question 11:

If A and B are acute angles such that sin A = cos B then (A + B) = ?
(a) 45°
(b) 60°
(c) 90°
(d) 180°