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