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Prove the following trigonometric identities:$ \sqrt{\frac{1-\cos A}{1+\cos A}}+\sqrt{\frac{1+\cos A}{1-\cos A}}=2 \operatorname{cosec} A $
To do:
We have to prove that \( \sqrt{\frac{1-\cos A}{1+\cos A}}+\sqrt{\frac{1+\cos A}{1-\cos A}}=2 \operatorname{cosec} A \).
Solution:
We know that,
$\sin ^{2} A+\cos^2 A=1$.......(i)
$\operatorname{cosec} A=\frac{1}{\sin A}$......(ii)
Therefore,
$\sqrt{\frac{1-\cos \mathrm{A}}{1+\cos \mathrm{A}}} +\sqrt{\frac{1+\cos \mathrm{A}}{1-\cos \mathrm{A}}}=\sqrt{\frac{(1-\cos A)(1-\cos A)}{(1+\cos A)(1-\cos A)}}+\sqrt{\frac{(1+\cos A)(1+\cos A)}{(1-\cos A)(1+\cos A)}}$ (Rationalizing the denominators)
$=\sqrt{\frac{(1-\cos A)^{2}}{1-\cos ^{2} A}}+\sqrt{\frac{(1+\cos A)^{2}}{1-\cos ^{2} A}}$
$=\sqrt{\frac{(1-\cos A)^{2}}{\sin ^{2} A}}+\sqrt{\frac{(1+\cos A)^{2}}{\sin ^{2} A}}$
$=\frac{1-\cos A}{\sin A}+\frac{1+\cos A}{\sin A}$
$=\frac{1-\cos A+1+\cos A}{\sin A}$
$=\frac{2}{\sin A}$
$=2 \operatorname{cosec} A$
Hence proved.
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