Differentiation — Maths STD 12 Science — Question

Differentiating w.r.t. x, we get

$\begin{aligned} f^{\prime}(x) & =\frac{d}{d x}\left(\tan ^{-1} x+\cot ^{-1} x\right) \\ & =\frac{d}{d x}\left(\tan ^{-1} x\right)+\frac{d}{d x}\left(\cot ^{-1} x\right) \\ & =\frac{1}{1+x^2}-\frac{1}{1+x^2}=0\end{aligned}$

Since, f'(x) = 0, f(x) is a constant function. Let f(x) = k. For any value of x, f(x) = k Let x = 0. Then f(0) = k ….(2) From (1), f(0) = tan-1(0) + cot-1(0)

$=0+\frac{\pi}{2}=\frac{\pi}{2}$

$\therefore k=\frac{\pi}{2} \quad \ldots[$ By (2)]

$\therefore f(x)=k=\frac{\pi}{2}$

Hence, $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2} . \quad \ldots$ [By (1)]