Let f(x) = tan^{-1}left(frac{1+cos x}{sin x}r | vedclass

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(B) First,simplify $f(x)$:\ $f(x) = 	an^{-1}\left(\frac{2\cos^2(x/2)}{2\sin(x/2)\cos(x/2)}ight) = 	an^{-1}(\cot(x/2)) = 	an^{-1}(	an(\pi/2 - x/

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$\pi x - \frac{x^2}{2} + C$

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(B) First,simplify $f(x)$:\ $f(x) = 	an^{-1}\left(\frac{2\cos^2(x/2)}{2\sin(x/2)\cos(x/2)}ight) = 	an^{-1}(\cot(x/2)) = 	an^{-1}(	an(\pi/2 - x/2)) = \frac{\pi}{2} - \frac{x}{2}$.\ Next,simplify $g(x)$:\ $g(x) = 	an^{-1}\left(\frac{2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)}ight) = 	an^{-1}(\cot(x/2)) = 	an^{-1}(	an(\pi/2 - x/2)) = \frac{\pi}{2} - \frac{x}{2}$.\ Now,calculate the sum:\ $f(x) + g(x) = (\frac{\pi}{2} - \frac{x}{2}) + (\frac{\pi}{2} - \frac{x}{2}) = \pi - x$.\ Finally,integrate the sum:\ $\int (\pi - x) \, dx = \pi x - \frac{x^2}{2} + C$.\ Thus,option $B$ is correct.

Let $f(x) = \tan^{-1}\left(\frac{1+\cos x}{\sin x}\right)$ and $g(x) = \tan^{-1}\left(\frac{\sin x}{1-\cos x}\right)$. Then,evaluate $\int (f(x) + g(x)) \, dx$.

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