Exercise 6.1 - Chapter 6 Trigonometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]
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On April 24, 2024, 11:35 AM
Ex $6.1$ : Chapter 6 - Trigonometry - 10th Maths Guide Samacheer Kalvi Solutions
Question 1.
Prove the following identities.
(i) $\cot \theta+\tan \theta=\sec \theta \operatorname{cosec} \theta$
(ii) $\tan ^4 \theta+\tan ^2 \theta=\sec ^4 \theta-\sec ^2 \theta$
Solution:
(i) L
$
\begin{aligned}
& \text { L.H.S }=\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta} \quad \begin{array}{r}
\text { Hint: } 1+\tan ^2 \theta=\sec ^2 \theta \\
\tan ^2 \theta=\sec ^2 \theta-1
\end{array} \\
& =\frac{\cos ^2 \theta+\sin ^2 \theta}{\sin \theta \cos \theta}=\frac{1}{\sin \theta \cos \theta} \\
& =\frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta}=\sec \theta \operatorname{cosec} \theta=\text { R.H.S }
\end{aligned}
$
(ii) L.H.S $=\tan ^2 \theta\left(\tan ^2 \theta+1\right)$
$
\begin{aligned}
& =\tan ^2 \theta\left(\sec ^2 \theta\right) \\
& =\left(\sec ^2 \theta-1\right)\left(\sec ^2 \theta\right) \\
& =\sec ^4 \theta-\sec ^2 \theta=\text { R.H.S }
\end{aligned}
$
Question 2.
Prove the following identities
(i) $\frac{1-\tan ^2 \theta}{\cot ^2 \theta-1}=\tan ^2 \theta$
(ii) $\frac{\cos \theta}{1+\sin \theta}=\sec \theta-\tan \theta$
Solution:
$
\begin{aligned}
& \text { L.H.S }=\frac{1-\tan ^2 \theta}{\cot ^2 \theta-1}=\frac{1-\left(\sec ^2 \theta-1\right)}{\left(\operatorname{cosec}^2 \theta-1\right)-1} \\
& \text { Hint: } 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta \\
& =\frac{\frac{1-\sin ^2 \theta}{\cos ^2 \theta}}{\frac{\cos ^2 \theta}{\sin ^2 \theta}-1}=\frac{\frac{\cos ^2 \theta-\sin ^2 \theta}{\cos ^2 \theta}}{\frac{\cos ^2 \theta-\sin ^2 \theta}{\sin ^2 \theta}} \\
& =\frac{\cos ^2 \theta-\sin ^2 \theta}{\cos ^2 \theta} \times \frac{\sin ^2 \theta}{\cos ^2 \theta-\sin ^2 \theta} \quad \begin{array}{c}
\text { Hint: } \\
\sin ^2 \theta+\cos ^2 \theta=1
\end{array} \\
& \cos ^2 \theta=1-\sin ^2 \theta \\
& =\frac{\sin ^2 \theta}{\cos ^2 \theta}=\tan ^2 \theta=\text { R.H.S } \\
& 2 \cos ^2 \theta=2-2 \sin ^2 \theta \\
&
\end{aligned}
$
(ii)
$
\begin{aligned}
& \text { L.H.S }=\frac{\cos \theta}{1+\sin \theta} \\
& \quad \begin{array}{r}
\text { Multiplying Numerator and } \\
\text { denominator by }(1-\sin \theta)
\end{array} \\
& =\frac{\cos \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}=\frac{\cos \theta-\cos \theta \sin \theta}{1-\sin ^2 \theta} \\
& =\frac{\cos \theta-\cos ^2 \theta \sin \theta}{\cos ^2 \theta}=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta} \\
& =\sec \theta-\tan ^2 \theta=\text { R.H.S }
\end{aligned}
$
Question 3.
Prove the following identities
(i) $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}=\sec \theta+\tan \theta$
(ii) $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=2 \sec \theta$
Solution:
(i) $\begin{aligned} \text { L.H.S } & =\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}} \\ = & \sqrt{\frac{(1+\sin \theta)^2}{1-\sin ^2 \theta}}=\sqrt{\frac{(1+\sin \theta)^2}{\cos ^2 \theta}} \\ = & \frac{1+\sin \theta}{\cos \theta}=\sec \theta+\tan \theta=\text { R.H.S }\end{aligned}$
$
\text { (ii) } \begin{aligned}
\text { L.H.S }=\sqrt{\frac{1+\sin \theta}{1-\sin \theta}} & =\sqrt{\frac{1+\sin \theta}{1-\sin \theta}} \times \frac{\sqrt{1+\sin \theta}}{1+\sin \theta} \\
=\sqrt{\frac{(1+\sin \theta)^2}{1-\sin ^2 \theta}} & =\frac{1+\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta} \\
& =\sec \theta+\tan \theta \ldots(1) \\
\sqrt{\frac{1-\sin \theta}{1+\sin \theta}} & =\sqrt{\frac{1-\sin \theta}{1+\sin \theta}} \times \frac{\sqrt{1-\sin \theta}}{1-\sin \theta} \\
& =\frac{1-\sin \theta}{\sqrt{1-\sin ^2 \theta}} \\
& =\frac{1-\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta} \\
= & \sec \theta-\tan \theta \quad \ldots \text { (2) }
\end{aligned}
$
$
\begin{aligned}
& (1)+(2) \Rightarrow \sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}} \\
& =\sec \theta+\tan \theta+\sec \theta-\tan \theta \\
& =2 \sec \theta=\text { R.H.S } \quad \text { Hence proved }
\end{aligned}
$
Question 4.
Prove the following identities
(i) $\sec ^6 \theta=\tan ^6 \theta+3 \tan ^2 \theta \sec ^2 \theta+1$
(ii) $(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2$
$
=1+(\sec \theta+\operatorname{cosec} \theta)^2
$
(i) L.H.S $=\sec ^6 \theta=\left(\sec ^2 \theta\right)^3=\left(1+\tan ^2 \theta\right)^3=\left(\tan ^2 \theta+1\right)^3$
$
(a+b)^3=a^3+3 a^2 b+3 a b^2+b^3
$
$
\begin{aligned}
& =\left(\tan ^2 \theta\right)^3+3\left(\tan ^2 \theta\right)^2 \times 1+3 \times \tan ^2 \theta \times 1^2+1 \\
& =\tan ^6 \theta+3 \tan ^2 \theta \times\left(\sec ^2 \theta-1\right)+3 \tan ^2 \theta+1 \\
& =\tan ^6 \theta+3 \tan ^2 \theta \sec ^2 \theta-3 \tan ^2 \theta+3 \tan ^2 \theta+1 \\
& =\tan ^6 \theta+3 \tan ^2 \theta \sec ^2 \theta+1=\text { R.H.S }
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) L.H.S }=(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2 \\
& =\sin ^2 \theta+2 \sin \theta \sec \theta+\sec ^2 \theta+\cos ^2 \theta+2 \cos \theta \operatorname{cosec} \theta+\operatorname{cosec}^2 \theta \\
& 1+\sec ^2 \theta+\operatorname{cosec}{ }^2 \theta+2 \sin \theta \sec \theta+2 \cos \theta \\
& \operatorname{cosec} \theta
\end{aligned}
$
$
\begin{aligned}
& 1+\sec ^2 \theta+\operatorname{cosec}^2 \theta+2\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right) \\
& 1+\sec ^2 \theta+\operatorname{cosec}^2 \theta+2\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}\right) \\
& 1+\sec ^2 \theta+\operatorname{cosec}^2 \theta+2 \times\left(\frac{1}{\sin \theta \cos \theta}\right) \\
& =1+\left(\sec ^2 \theta+\operatorname{cosec}^2 \theta+2 \sec \theta \operatorname{cosec} \theta\right) \\
& =1+(\sec \theta+\operatorname{cosec} \theta)^2=\text { R.H.S }
\end{aligned}
$
Question 5.
Prove the following identities
(i) $\sec ^4 \theta\left(1-\sin ^4 \theta\right)-2 \tan ^2 \theta=1$
(ii) $\frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta}=\frac{\operatorname{cosec} \theta-1}{\operatorname{cosec} \theta+1}$
Solution:
L.H.S $=\sec ^4 \theta\left(1-\sin ^4 \theta\right)-2 \tan ^2 \theta$
$
\begin{aligned}
& \frac{1-\sin ^4 \theta}{\cos ^4 \theta}-2 \tan ^2 \theta \\
= & \sec ^4 \theta-\tan ^4 \theta-2 \tan ^2 \theta \\
= & \frac{1}{\cos ^4 \theta}-\frac{\sin ^4 \theta}{\cos ^4 \theta}-\frac{2 \sin ^2 \theta}{\cos ^2 \theta} \\
= & \frac{1-\sin ^4 \theta-2 \sin ^2 \theta \cos ^2 \theta}{\cos ^4 \theta} \\
= & 1+\frac{\cos ^4 \theta-\cos ^4 \theta-\sin ^4 \theta-2 \sin ^2 \theta \cos ^2 \theta}{\cos ^4 \theta} \\
= & \frac{1+\cos ^4 \theta-\left[\sin ^4 \theta+2 \sin ^2 \theta \cos ^2 \theta+\cos ^4 \theta\right]}{\cos ^4 \theta} \\
= & \frac{1+\cos ^4 \theta-\left(\sin ^2 \theta+\cos ^2 \theta\right)^2}{\cos ^4 \theta} \\
= & \frac{1+\cos ^4 \theta-1}{\cos ^4 \theta} \\
= & \frac{\cos ^4 \theta}{\cos ^4 \theta}=1=\text { R.H.S }
\end{aligned}
$
$\begin{aligned}
\frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta} & =\frac{\frac{\cos \theta}{\sin \theta}-\cos \theta}{\frac{\cos \theta}{\sin \theta}+\cos \theta} \\
& =\frac{\cos \theta\left(\frac{1}{\sin \theta}-1\right)}{\cos \theta\left(\frac{1}{\sin \theta}+1\right)} \\
& =\frac{\operatorname{cosec} \theta-1}{\operatorname{cosec} \theta+1}=\text { R.H.S }
\end{aligned}$
Question 6.
Prove the following identities
(i) $\frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B}=0$
(ii) $\frac{\sin ^3 A+\cos ^3 A}{\sin A+\cos A}+\frac{\sin ^3 A-\cos ^3 A}{\sin A-\cos A}=2$
Solution:
$
\begin{aligned}
& \text { (i) } \mathrm{LHS}=\frac{\sin \mathrm{A}-\sin \mathrm{B}}{\cos \mathrm{A}+\cos \mathrm{B}}+\frac{\cos \mathrm{A}-\cos \mathrm{B}}{\sin \mathrm{A}+\sin \mathrm{B}} \\
& =\frac{(\sin A-\sin B)(\sin A+\sin B)+(\cos A+\cos B)(\cos A-\cos B)}{(\cos A+\cos B)(\sin A+\sin B)} \\
& =\frac{\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}+\cos ^2 \mathrm{~A}-\cos ^2 \mathrm{~B}}{(\cos \mathrm{A}+\cos \mathrm{B})(\sin \mathrm{A}+\sin \mathrm{B})} \\
& =\frac{\left(\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}\right)-\left(\sin ^2 \mathrm{~B}+\cos ^2 \mathrm{~B}\right)}{(\cos \mathrm{A}+\cos \mathrm{B})(\sin \mathrm{A}+\sin \mathrm{B})} \\
& =\frac{1-1}{(\cos A+\cos B)(\sin A+\sin B)}=0 \\
& =\text { R.H.S } \\
&
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) } \frac{\sin ^3 \mathrm{~A}+\cos ^3 \mathrm{~A}}{\sin \mathrm{A}+\cos \mathrm{A}}+\frac{\sin ^3 \mathrm{~A}-\cos ^3 \mathrm{~A}}{\sin \mathrm{A}-\cos \mathrm{A}}=2 \\
& (a+b)^3=a^3+3 a^2 b+3 a b^2+b^3 \\
& \therefore a^3+b^3=(a+b)^3-3 a^2 b-3 a b^2 \\
& (a-b)^3=a^3-3 a^2 b+3 a b^2-b^3 \\
& \therefore a^3-b^3 \quad=(a-b)^3+3 a^2 b-3 a b^2 \\
& \sin ^3 \mathrm{~A}+\cos ^3 \mathrm{~A}=(\sin \mathrm{A}+\cos \mathrm{A})^3-3 \sin ^2 \mathrm{~A} \cos \mathrm{A} \\
& -3 \sin A \cos ^2 A \\
& (\sin \mathrm{A}+\cos \mathrm{A})^3-3 \sin \mathrm{A} \cos \mathrm{A} \\
& (\sin \mathrm{A}+\cos \mathrm{A}) \\
& \sin ^3 \mathrm{~A}-\cos ^3 \mathrm{~A}=(\sin \mathrm{A}-\cos \mathrm{A})^3+3 \sin \mathrm{A} \cos \mathrm{A} \\
& (\sin A-\cos A) \\
&
\end{aligned}
$
Substituting (1) and (2) in LHS, we get
$
\begin{aligned}
\text { L.H.S } & =(\sin A+\cos A)^2-3 \sin A \cos A+ \\
= & \underbrace{\sin ^2 A+\cos ^2 A}_1+2 \sin A \cos A \\
& +\underbrace{\sin ^2 A+\cos ^2 A}_1-2 \sin A \cos A \\
= & 1+1=2=\text { R.H.S }
\end{aligned}
$
Question 7.
(i) If $\sin \theta+\cos \theta=\sqrt{3}$, then prove that $\tan \theta+\cot \theta=1$.
(ii) If $\sqrt{3} \sin \theta-\cos \theta=0$, then show that $\tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}$.
Solution:
(i) Given $\sin \theta+\cos \theta=\sqrt{3}$
Squaring both sides we get
$
\begin{aligned}
& \sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta=3 \\
& \therefore \sin \theta \cos \theta=1
\end{aligned}
$
use (1) to prove, that $\tan \theta+\cot \theta=1$ as follows
$
\begin{aligned}
\text { L.H.S } & =\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta} \\
& =\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta} \\
& =\frac{1}{\cos \theta \sin \theta}=\frac{1}{1}=1 \\
& =\text { R.H.S. }
\end{aligned}
$
(ii) $\sqrt{3} \sin \theta-\cos \theta=0 \Rightarrow \sqrt{3} \sin \theta=\cos \theta$
$
\begin{aligned}
\therefore \quad \tan \theta & =\frac{1}{\sqrt{3}} \Rightarrow \theta=30^{\circ} \\
\text { L.H.S } & =\tan 3 \theta \\
& =\tan (3 \times 30)=\tan 90=\infty \\
\text { R.H.S } & =\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}
\end{aligned}
$
Substituting $\tan \theta=\frac{1}{\sqrt{3}}$ we get
Question 8.
(i) If $\frac{\cos \alpha}{\cos \beta}=m$ and $\frac{\cos \alpha}{\sin \beta}=n$ then prove that $\left(m^2+n^2\right) \cos ^2 \beta=n^2$
(ii) If $\cot \theta+\tan \theta=x$ and $\sec \theta-\cos \theta=y$ then prove that $\left(x^2 y\right)^{\frac{2}{3}}-\left(x y^2\right)^{\frac{2}{3}}=1$
Solution:
(i) LHS:
$
\begin{aligned}
& \left(m^2+n^2\right) \cos ^2 \beta=\left(\frac{\cos ^2 \alpha}{\cos ^2 \beta}+\frac{\cos ^2 \alpha}{\sin ^2 \beta}\right) \times \cos ^2 \beta \\
& =\left(\frac{\sin ^2 \beta \cos ^2 \alpha+\cos ^2 \alpha \cos ^2 \beta}{\cos ^2 \beta \sin ^2 \beta}\right) \times \cos ^2 \beta \\
& =\cos ^2 \alpha+\frac{\cos ^2 \alpha\left(1-\sin ^2 \beta\right)}{\sin ^2 \beta} \\
& =\operatorname{sos}^2 \alpha+\frac{\cos ^2 \alpha}{\sin ^2 \beta}-\cos ^2 \alpha \\
& =\left(\frac{\cos \alpha}{\sin \beta}\right)^2=n^2=\text { R.H.S }
\end{aligned}
$
$\begin{aligned}
& \text { (ii) } x=\cot \theta+\tan \theta=\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta} \\
& =\frac{\cos ^2 \theta+\sin ^2 \theta}{\cos \theta \sin \theta}=\frac{1}{\sin \theta \cos \theta} \\
& y=\sec \theta-\cos \theta=\frac{1}{\cos \theta}-\cos \theta \\
& =\frac{1-\cos ^2 \theta}{\cos \theta} \\
& =\frac{\sin ^2 \theta}{\cos \theta} \\
& \therefore x^2 y=\frac{1}{\sin ^2 \theta \cos ^2 \theta} \times \frac{\sin ^2 \theta}{\cos \theta} \\
& =\frac{1}{\cos ^3 \theta} \\
& x y^2=\frac{1}{\sin \theta \cos \theta} \times \frac{\sin ^4 \theta}{\cos ^2 \theta} \\
& =\frac{\sin ^3 \theta}{\cos ^3 \theta} \\
& \left(x^2 y\right)^{\frac{2}{3}}=\left(\frac{1}{\cos ^3 \theta}\right)^{\frac{2}{3}}=\frac{1}{\cos ^2 \theta} \\
& \left(x y^2\right)^{\frac{2}{3}}=\left(\frac{\sin ^3 \theta}{\cos ^3 \theta}\right)^{\frac{2}{3}}=\frac{\sin ^2 \theta}{\cos ^2 \theta} \\
& \text { LHS }=\frac{1}{\cos ^2 \theta}-\frac{\sin ^2 \theta}{\cos ^2 \theta}=\frac{1-\sin ^2 \theta}{\cos ^2 \theta} \\
& =\frac{\cos ^2 \theta}{\cos ^2 \theta}=1=\text { R.H.S } \\
&
\end{aligned}$
Question 9.
(i) If $\sin \theta+\cos \theta=p$ and $\sec \theta+\operatorname{cosec} \theta=q$ then prove that $q\left(p^2-1\right)=2 p$
(ii) If $\sin \theta\left(1+\sin ^2 \theta\right)=\cos ^2 \theta$, then prove that $\cos ^6 \theta-4 \cos ^4 \theta+8 \cos ^2 \theta=4$ Solution:
$
\begin{aligned}
& \text { (i) } p=\sin \theta+\cos \theta \\
& p^2=\underbrace{\sin ^2 \theta+\cos ^2 \theta}_1+2 \sin \theta \cos \theta \\
& p^2-1=2 \sin \theta \cos \theta \\
& q=\sec \theta+\operatorname{cosec} \theta=\frac{1}{\cos \theta}+\frac{1}{\sin \theta} \\
& =\quad=\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta} \\
& \therefore \text { L.H.S } q\left(p^2-1\right)=\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta} \times 2 \sin \theta \cos \theta \\
& =2(\sin \theta+\cos \theta) \\
& =2 p=\mathrm{R} . \mathrm{H} \cdot \mathrm{S} \\
&
\end{aligned}
$
(ii) Given $\sin \theta\left(1+\sin ^2 \theta\right)=\cos ^2 \theta$
Sustitute $\sin ^2 \theta=1-\cos ^2 \theta$ and take $\cos \theta=\mathrm{c}$
squaring (1) on bothsides we get
$
\begin{aligned}
& \sin ^2 \theta\left(1+\sin ^2 \theta\right)^2=\cos ^4 \theta \\
& \left(1-c^2\right)\left(1+1-c^2\right)=c^4 \\
& \left(1-c^2\right)\left(2-c^2\right)^2=c^4 \\
& \left(1-c^2\right)\left(4+c^4-4 c^2\right)=c^4 \\
& 4+c^4-4 c^2-4 c^2-c^6+4 c^4=c^4 \\
& -c^6+4 c^4-8 c^2=-4 \\
& c^6-4 c^4+8 c^2=-4
\end{aligned}
$
ie $\cos 6 \theta-4 \cos 4 \theta+8 \cos ^2 \theta=4=$ RHS
$\therefore$ Hence proved
Question 10.
If $\frac{\cos \theta}{1+\sin \theta}=\frac{1}{a}$ then prove that $\frac{a^2-1}{a^2+1}=\sin \theta$
Solution:
$
\begin{aligned}
a^2 & =\frac{(1+\sin \theta)^2}{\cos ^2 \theta}=\frac{1+\sin ^2 \theta+2 \sin \theta}{\cos ^2 \theta} \\
\therefore a^2-1 & =\frac{\sin ^2 \theta+2 \sin \theta+1-\cos ^2 \theta}{\cos ^2 \theta} \\
& =\frac{\sin ^2 \theta+2 \sin \theta+\sin ^2 \theta}{\cos ^2 \theta} \\
& =\frac{2 \sin ^2 \theta+2 \sin \theta}{\cos ^2 \theta} \\
a^2+1 & =\frac{\sin ^2 \theta+2 \sin \theta+1+\cos ^2 \theta}{\cos ^2 \theta} \\
& =\frac{1+2 \sin \theta+1}{\cos ^2 \theta}=\frac{2+2 \sin \theta}{\cos ^2 \theta} \\
\therefore \text { L.H.S } \frac{a^2-1}{a^2+1} & =\frac{2 \sin \theta+2 \sin \theta}{2 \sin \theta+2} \\
& =\frac{2\sin \theta(\sin \theta+1)}{2(\sin \theta+1)} \\
& =\sin \theta=\text { R.H.S }
\end{aligned}
$
Also Read : Exercise-7.1-Chapter-7-Mensuration-10th-Maths-Guide-Samacheer-Kalvi-Solutions
