The solution of the equation tan ^{-1}(1+x)+t | vedclass

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(B) Given the equation: $	an ^{-1}(1+x)+	an ^{-1}(1-x)=\frac{\pi}{2}$ We know that $	an ^{-1}(A) + \cot ^{-1}(A) = \frac{\pi}{2}$,so $\frac{\pi}{2}

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$x=0$

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(B) Given the equation: $	an ^{-1}(1+x)+	an ^{-1}(1-x)=\frac{\pi}{2}$ We know that $	an ^{-1}(A) + \cot ^{-1}(A) = \frac{\pi}{2}$,so $\frac{\pi}{2} - 	an ^{-1}(1-x) = \cot ^{-1}(1-x)$. Substituting this into the equation: $	an ^{-1}(1+x) = \cot ^{-1}(1-x)$. Using the identity $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$,we get: $	an ^{-1}(1+x) = 	an ^{-1}(\frac{1}{1-x})$. Equating the arguments: $1+x = \frac{1}{1-x}$. This simplifies to: $(1+x)(1-x) = 1$. $1 - x^2 = 1$. $-x^2 = 0$,which implies $x = 0$.

The solution of the equation $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ is

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