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Trigonometric Identities — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities5 Marks
Question
Prove the following identities:
$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}+\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}=\frac{2}{\big(\sin^2\theta-\cos^2\theta\big)}=\frac{2}{\big(2\sin^2\theta-1\big)}$
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Answer

$\text{LHS}=\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}+\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}$
$=\frac{(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2}{(\sin\theta-\cos\theta)(\sin\theta+\cos\theta)}$
$=\frac{\big(\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta\big)+\big(\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta\big)}{\sin^2\theta-\cos^2\theta}$
$=\frac{(1+2\sin\theta\cos\theta)+(1-2\sin\theta\cos\theta)}{\sin^2\theta-\cos^2\theta}$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=\frac{2}{\sin^2\theta-\cos^2\theta}$
Also, $\frac{2}{\sin^2\theta-\cos^2\theta}=\frac{2}{\sin^2\theta-\big(1-\sin^2\theta\big)}$
$=\frac{2}{2\sin^2\theta-1}$
$=\text{R.H.S.}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$

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