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Trigonometric Identities — Maths STD 10 — Question

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

$\text{LHS}=\Bigg\{\frac{1}{\big(\sec^2\theta-\cos^2\theta\big)}+\frac{1}{\big(\text{cosec}^2\theta-\sin^2\theta\big)}\Bigg\}\big(\sin^2\theta\cos^2\theta\big)$
$=\Bigg\{\frac{1}{\frac{1}{\cos^2\theta}-\cos^2\theta}+\frac{1}{\frac{1}{\sin^2\theta}-\sin^2\theta}\Bigg\}\times\sin^2\theta\cos^2\theta$
$=\Bigg\{\frac{\sin^2\theta\cos^2\times\cos^2\theta}{\big(1-\cos^4\theta\big)}+\frac{\sin^4\theta\cos^2\theta\sin^2\theta}{1-\sin^4\theta}\Bigg\}$
$=\Bigg\{\frac{\sin^2\theta\times\cos^4\theta}{\big(1-\cos^2\theta\big)\big(1-\cos^2\theta\big)}+\frac{\sin^4\theta\cos^2\theta}{1-\sin^2\theta\big(1+\sin^2\theta\big)}\Bigg\}$
$=\Bigg\{\frac{\cos^4\theta}{\big(1+\cos^2\theta\big)}+\frac{\sin^4\theta}{1+\sin^2\theta}\Bigg\}$
$=\frac{\cos^4\theta+\cos^4\theta\sin^2\theta+\sin^4\theta+\sin^4\theta\cos^2\theta}{\big(1+\cos^2\theta\big)\big(1+\sin^2\theta\big)}$
$=\frac{\cos^4\theta+\sin^4\theta+\cos^2\theta\sin^2\theta\big(\cos^2\theta+\sin^2\theta\big)}{1+\sin^2\theta+\cos^2\theta+\cos^2\theta+\sin^2\theta}$ $\Big[\because\text{a}^2+\text{b}^2=(\text{a}+\text{b})^2-2\text{ab}\Big]$
$=\frac{1-\cos^2\theta\sin^2\theta}{2+\cos^2\theta\sin^2\theta}$
$=\text{R.H.S}$
$\therefore\ \text{R.H.S.}=\text{L.H.S.}$

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