If 0 leq A leq frac{pi}{4}, then tan ^{-1}lef | vedclass

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(C) Let $S = 	an ^{-1}\left(\frac{1}{2} 	an 2 Aight)+	an ^{-1}(\cot A)+	an ^{-1}(\cot ^{3} A)$.Using the identity $	an ^{-1} x + 	an ^{-1} y =

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(C) Let $S = 	an ^{-1}\left(\frac{1}{2} 	an 2 Aight)+	an ^{-1}(\cot A)+	an ^{-1}(\cot ^{3} A)$.Using the identity $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$, we have:$	an ^{-1}(\cot A) + 	an ^{-1}(\cot ^{3} A) = 	an ^{-1}\left(\frac{\cot A + \cot ^{3} A}{1 - \cot A \cdot \cot ^{3} A}ight) = 	an ^{-1}\left(\frac{\cot A(1 + \cot ^{2} A)}{1 - \cot ^{4} A}ight)$.Since $1 - \cot ^{4} A = (1 - \cot ^{2} A)(1 + \cot ^{2} A)$, the expression simplifies to:$	an ^{-1}\left(\frac{\cot A}{1 - \cot ^{2} A}ight) = 	an ^{-1}\left(\frac{1/	an A}{1 - 1/	an ^{2} A}ight) = 	an ^{-1}\left(\frac{	an A}{	an ^{2} A - 1}ight) = -	an ^{-1}\left(\frac{	an A}{1 - 	an ^{2} A}ight)$.Now, $\frac{1}{2} 	an 2 A = \frac{1}{2} \cdot \frac{2 	an A}{1 - 	an ^{2} A} = \frac{	an A}{1 - 	an ^{2} A}$.Thus, $S = 	an ^{-1}\left(\frac{	an A}{1 - 	an ^{2} A}ight) - 	an ^{-1}\left(\frac{	an A}{1 - 	an ^{2} A}ight) = 0$.

If $0 \leq A \leq \frac{\pi}{4},$ then $\tan ^{-1}\left(\frac{1}{2} \tan 2 A\right)+\tan ^{-1}(\cot A)+\tan ^{-1}(\cot ^{3} A)$ is equal to

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