Question
If $\sec A+\tan A=p$, show that: $\sin A =\frac{p^2-1}{p^2+1}$
✓
Answer
$\frac{p^2-1}{p^2+1}$
$=\frac{(\sec A +\tan A )^2-1}{(\sec A +\tan A )^2+1}$
$=\frac{\sec ^2 A +\tan ^2 A +2 \tan A \sec A -1}{\sec ^2 A +\tan ^2 A +2 \tan A \sec A +1}$
$=\frac{\tan ^2 A +\tan ^2 A +2 \tan A \sec A }{\sec ^2 A +\sec ^2 A +2 \tan A \sec A }$
$=\frac{2 \tan ^2 A +2 \tan A \sec A }{2 \sec ^2 A +2 \tan A \sec A }$
$=\frac{2 \tan A (\tan A +\sec A )}{2 \sec A (\tan A +\sec A )}$
$=\sin A $
$=\frac{(\sec A +\tan A )^2-1}{(\sec A +\tan A )^2+1}$
$=\frac{\sec ^2 A +\tan ^2 A +2 \tan A \sec A -1}{\sec ^2 A +\tan ^2 A +2 \tan A \sec A +1}$
$=\frac{\tan ^2 A +\tan ^2 A +2 \tan A \sec A }{\sec ^2 A +\sec ^2 A +2 \tan A \sec A }$
$=\frac{2 \tan ^2 A +2 \tan A \sec A }{2 \sec ^2 A +2 \tan A \sec A }$
$=\frac{2 \tan A (\tan A +\sec A )}{2 \sec A (\tan A +\sec A )}$
$=\sin A $
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