Question
Prove the following.
$\sec ^2 \theta+\operatorname{cosec}^2 \theta=\sec ^2 \theta \times \operatorname{cosec}^2 \theta$
$\sec ^2 \theta+\operatorname{cosec}^2 \theta=\sec ^2 \theta \times \operatorname{cosec}^2 \theta$
✓
Answer
Taking LHS
$=\sec ^2 \theta + \operatorname{cosec}^2 \theta$
$=\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}$
$=\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos ^2 \theta \sin ^2 \theta}$
$ =\frac{1}{\cos ^2 \theta \sin ^2 \theta}\left[\sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\sec ^2 \theta \times \operatorname{cosec}^2 \theta$
$=\text { RHS }$
Proved!
$=\sec ^2 \theta + \operatorname{cosec}^2 \theta$
$=\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}$
$=\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos ^2 \theta \sin ^2 \theta}$
$ =\frac{1}{\cos ^2 \theta \sin ^2 \theta}\left[\sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\sec ^2 \theta \times \operatorname{cosec}^2 \theta$
$=\text { RHS }$
Proved!
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