Total number of solutions of the equation $\cos^{-1}(\frac{1-x^2}{1+x^2})=\sin^{-1}x$ is:

Two

$\cos^{-1}(\frac{1-x^2}{1+x^2})=\sin^{-1}x⇒\sin^{-1}(\frac{2x}{1+x^2})=\sin^{-1}(x)⇒\frac{2x}{1+x^2}=x$

$⇒2=1+x^2∵x=±1,x=0$

$⇒-1≤\frac{1-x^2}{1+x^2}≤1⇒-1-x^2≤1-x^2≤1+x^2⇒2x^2≥0⇒x∈R$

Also, -1 ≤ x ≤ 1 (Note that x = -1 does not satisfy the original equation) ⇒ Only 2 solutions