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If $\sec A+\tan A=p$, then what is the value of $\sin A$ ? $[2008-\mathrm{II}]$
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Verified Answer
The correct answer is:
$\frac{p^{2}-1}{p^{2}+1}$
$\sec A+\tan A=P$
$\Rightarrow \frac{1}{\cos A}+\frac{\sin A}{\cos A}=p$
$\Rightarrow \frac{1+\sin A}{\cos A}=p$
$\Rightarrow \frac{(1+\sin A)^{2}}{\cos ^{2} A}=p^{2}$
$\Rightarrow \frac{(1+\sin A)^{2}}{1-\sin ^{2} A}=p^{2}$
$\Rightarrow \frac{(1+\sin A)^{2}}{(1+\sin A)(1-\sin A)}=p^{2}$
$\Rightarrow \frac{1}{\cos A}+\frac{\sin A}{\cos A}=p$
$\Rightarrow \frac{1+\sin A}{\cos A}=p$
$\Rightarrow \frac{(1+\sin A)^{2}}{\cos ^{2} A}=p^{2}$
$\Rightarrow \frac{(1+\sin A)^{2}}{1-\sin ^{2} A}=p^{2}$
$\Rightarrow \frac{(1+\sin A)^{2}}{(1+\sin A)(1-\sin A)}=p^{2}$
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