## Trigonometry (11th Edition) Clone

$x = 0$
We can use the following identity: $cos^{-1}~a+cos^{-1}~b = cos^{-1}[ab-\sqrt{(1-a^2)(1-b^2)}]$ We can find the solution: $cos^{-1}~x+tan^{-1}~x = \frac{\pi}{2}$ $cos^{-1}~x+cos^{-1}~\frac{1}{\sqrt{x^2+1}} = \frac{\pi}{2}$ $cos^{-1}~[\frac{x}{\sqrt{x^2+1}}-\sqrt{(1-x^2)(1-\frac{1}{x^2+1})}~] = \frac{\pi}{2}$ $cos^{-1}~[\frac{x}{\sqrt{x^2+1}}-\sqrt{(1-x^2)(\frac{x^2}{x^2+1})}~] = \frac{\pi}{2}$ $cos^{-1}~(\frac{x}{\sqrt{x^2+1}}-\frac{x~\sqrt{1-x^2}}{\sqrt{x^2+1}}) = \frac{\pi}{2}$ $\frac{x}{\sqrt{x^2+1}}-\frac{x~\sqrt{1-x^2}}{\sqrt{x^2+1}} = 0$ $\frac{x-x~\sqrt{1-x^2}}{\sqrt{x^2+1}} = 0$ $x-x\sqrt{1-x^2} = 0$ $x(1-\sqrt{1-x^2}) = 0$ $x = 0$