Question:
Mark $(\sqrt{)}$ against the correct answer in the following:
If $y=\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}$ then $\frac{d y}{d x}=$
A. $\frac{1}{2}$
B. $\frac{-1}{2}$
C. $\frac{1}{\left(1+x^{2}\right)}$
D. none of these
Solution:
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