Question
Prove that $\sec^2\theta + cosec^2\theta = \sec^2\theta \times cosec^2\theta $
✓
Answer
$\text { L.H.S }=\sec ^2 \theta+\operatorname{cosec}^2 \theta$
$=\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}$
$=\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos ^2 \theta \cdot \sin ^2 \theta}$
$=\frac{1}{\cos ^2 \theta \cdot \sin ^2 \theta} \quad \ldots \ldots\left[\because \cdot \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\frac{1}{\cos ^2 \theta} \times \frac{1}{\sin ^2 \theta}$
$=\sec ^2 \theta \times \operatorname{cosec}^2 \theta$
$=\text { R.H.S }$
$\therefore \sec ^2 \theta+\operatorname{cosec}^2 \theta=\sec ^2 \theta \times \operatorname{cosec}^2 \theta$
$=\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}$
$=\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos ^2 \theta \cdot \sin ^2 \theta}$
$=\frac{1}{\cos ^2 \theta \cdot \sin ^2 \theta} \quad \ldots \ldots\left[\because \cdot \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\frac{1}{\cos ^2 \theta} \times \frac{1}{\sin ^2 \theta}$
$=\sec ^2 \theta \times \operatorname{cosec}^2 \theta$
$=\text { R.H.S }$
$\therefore \sec ^2 \theta+\operatorname{cosec}^2 \theta=\sec ^2 \theta \times \operatorname{cosec}^2 \theta$
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