Question
Prove the following identities:
$\frac{\cot^2\theta(\sec\theta-1)}{(1+\sin\theta)}+\frac{\sec^2\theta(\sin\theta-1)}{(1+\sec\theta)}=0$
$\frac{\cot^2\theta(\sec\theta-1)}{(1+\sin\theta)}+\frac{\sec^2\theta(\sin\theta-1)}{(1+\sec\theta)}=0$
✓
Answer
$\text{LHS}=\frac{\cot^2\theta(\sec\theta-1)}{(1+\sin\theta)}+\frac{\sec^2\theta(\sin\theta-1)}{(1+\sec\theta)}$
$=\frac{\cot^2\theta(\sec\theta-1)(1+\sec\theta)+\sec^2\theta(\sin\theta-1)(1+\sin\theta)}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\big(\sec^2\theta-1\big)+\sec^2\theta\big(\sin^2\theta-1\big)}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\tan^2\theta+\sec^2\theta\big(-\cos^2\theta\big)}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\tan^2\theta-\sec^2\theta\cos^2\theta}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\times\frac{1}{\cot^2\theta}-\sec^2\theta\times\frac{1}{\sec^2\theta}}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{1-1}{(1+\sin\theta)(1+\sec\theta)}$
$=0$
$\therefore\ \text{R.H.S.}=\text{L.H.S.}$
$=\frac{\cot^2\theta(\sec\theta-1)(1+\sec\theta)+\sec^2\theta(\sin\theta-1)(1+\sin\theta)}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\big(\sec^2\theta-1\big)+\sec^2\theta\big(\sin^2\theta-1\big)}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\tan^2\theta+\sec^2\theta\big(-\cos^2\theta\big)}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\tan^2\theta-\sec^2\theta\cos^2\theta}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{\cot^2\theta\times\frac{1}{\cot^2\theta}-\sec^2\theta\times\frac{1}{\sec^2\theta}}{(1+\sin\theta)(1+\sec\theta)}$
$=\frac{1-1}{(1+\sin\theta)(1+\sec\theta)}$
$=0$
$\therefore\ \text{R.H.S.}=\text{L.H.S.}$
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