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Question

$\sqrt{\frac{1+\sin\text{A}}{1-\sin\text{A}}}=\sec\text{A}+\tan\text{A}$

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Answer

$\text{L.H.S}=\sqrt{\frac{1+\sin\text{A}}{1-\sin\text{A}}}$
$=\sqrt{\frac{(1+\sin\text{A})(1+\sin\text{A})}{(1-\sin\text{A})(1+\sin\text{A})}}$
$=\sqrt{\frac{(1+\sin\text{A})^2}{1-\sin^2\text{A}}}=\sqrt{\frac{(1+\sin\text{A})^2}{\cos^2\text{A}}}$
$\frac{1+\sin\text{A}}{\cos\text{A}}=\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}$
$=\sec\text{A}+\tan\text{A}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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