Exercise 3.1-Additional Question - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Equations
Question 1.
Prove that $(\sec \theta+\cos \theta)(\sec \theta-\cos \theta)=\tan ^2 \theta+\sin ^2 \theta$
Solution:
$
\begin{aligned}
& (\sec \theta+\cos \theta)(\sec \theta-\cos \theta)=\sec ^2 \theta-\cos ^2 \theta \\
& =\left(1+\tan ^2 \theta\right)-\left(1-\sin ^2 \theta\right) \\
& =\tan ^2 \theta+\sin ^2 \theta=\text { RHS }
\end{aligned}
$
Question 2 .
If $\tan \theta+\sin \theta=p, \tan \theta-\sin \theta=q$ and $p>q$ then show that $p^2-q^2=4 \sqrt{p q}$.
Solution:
We know that $\sec ^2 \theta-\tan ^2 \theta=1$
$(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1$

$
\begin{aligned}
& \sec \theta-\tan \theta=\frac{1}{\sec \theta+\tan \theta} \\
& \sec \theta-\tan \theta=\frac{1}{x} \rightarrow(1) \\
& \sec \theta+\tan \theta=x \rightarrow(2) \\
& \text { (1) }+(2) \Rightarrow 2 \sec \theta=x+1 / x \\
& \text { (1) }-(2) \Rightarrow 2 \tan \theta=x-1 / x \\
& \frac{(4)}{(3)} \Rightarrow \frac{\tan \theta}{\sec \theta}=\frac{x-\frac{1}{x}}{x+\frac{1}{x}}=\frac{x^2-1}{x^2+1} \\
& \frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta}}=\frac{x^2-1}{x^2+1} \Rightarrow \sin \theta=\frac{x^2-1}{x^2+1} \\
&
\end{aligned}
$
$
[\sec \theta+\tan \theta=x]
$
Question 3.
If $\tan \theta+\sin \theta=p, \tan \theta-\sin \theta=q$ and $p>q$ then show that $p^2-q^2=4 \sqrt{p q}$.

Solution:
$
\begin{aligned}
\text { LHS }=p^2-q^2 & =(\tan \theta+\sin \theta)^2-(\tan \theta-\sin \theta)^2 \\
& =2 \tan \theta \cdot 2 \sin \theta=4 \tan \theta \sin \theta \\
p q & =(\tan \theta+\sin \theta)(\tan \theta-\sin \theta) \\
& =\tan ^2 \theta-\sin ^2 \theta=\frac{\sin ^2 \theta}{\cos ^2 \theta}-\sin ^2 \theta \\
& =\sin ^2 \theta\left(\frac{1}{\cos ^2 \theta}-1\right)=\sin ^2 \theta\left(\sec ^2 \theta-1\right) \\
& =\sin ^2 \theta \tan ^2 \theta \\
\therefore \text { RHS } & =4 \sqrt{p q}=4 \sqrt{\sin ^2 \theta \tan ^2 \theta}=4 \tan \theta \sin \theta \\
\text { LHS } & =\text { RHS }
\end{aligned}
$
From (1) and (2) $p^2-q^2=4 \sqrt{p q}$
Question 4.
If $\cos \theta+\sin \theta=\sqrt{2} \cos \theta$, show that $\cos \theta-\sin \theta=\sqrt{2} \sin \theta$.
Solution:
$
\cos \theta+\sin \theta=\sqrt{2} \cos \theta
$
Squaring on both sides $(\cos \theta+\sin \theta)^2=2 \cos ^2 \theta$
$
\begin{aligned}
\cos ^2 \theta+\sin ^2 \theta+2 \cos \theta \sin \theta & =2 \cos ^2 \theta \\
\sin ^2 \theta & =2 \cos ^2 \theta-\cos ^2 \theta-2 \sin \theta \cos \theta \\
& =\cos ^2 \theta-2 \sin \theta \cos \theta
\end{aligned}
$
Adding $\sin ^2 \theta$ on both sides we get, $2 \sin ^2 \theta=\cos ^2 \theta-2 \sin \theta \cos \theta+\sin ^2 \theta$
$
=(\cos \theta-\sin \theta)^2
$
$
\Rightarrow \quad \cos \theta-\sin \theta=\sqrt{2} \sin \theta
$
Question 5 .
Prove that $(1+\tan A+\sec A)(1+\cot A-\operatorname{cosec} A)=2$
Solution:
$
\text { LHS }=(1+\tan A+\sec A)(1+\cot A-\operatorname{cosec} A)
$

$\begin{aligned}
\text { LHS } & =(1+\tan \mathrm{A}+\sec \mathrm{A})(1+\cot \mathrm{A}-\operatorname{cosec} \mathrm{A}) \\
& =\left(1+\frac{\sin \mathrm{A}}{\cos \mathrm{A}}+\frac{1}{\cos \mathrm{A}}\right)\left(1+\frac{\cos \mathrm{A}}{\sin \mathrm{A}}-\frac{1}{\sin \mathrm{A}}\right) \\
& =\left(\frac{\cos \mathrm{A}+\sin \mathrm{A}+1}{\cos \mathrm{A}}\right)\left(\frac{\sin \mathrm{A}+\cos \mathrm{A}-1}{\sin \mathrm{A}}\right) \\
& =\frac{(\cos \mathrm{A}+\sin \mathrm{A})^2-1}{\sin \mathrm{A} \cos \mathrm{A}} \frac{\cos ^2 \mathrm{~A}+\sin ^2 \mathrm{~A}+2 \cos \mathrm{A} \sin \mathrm{A}-1}{\sin \mathrm{A} \cos \mathrm{A}} \\
& =\frac{2 \sin \mathrm{A} \cos \mathrm{A}}{\sin \mathrm{A} \cos \mathrm{A}}=2=\text { RHS }
\end{aligned}$