If $ \sin^{-1}(x)+\tan^{-1}(x) = \frac{\pi}{2}$, find $2x^2+1$

Question

If $ \sin^{-1}(x)+\tan^{-1}(x) = \frac{\pi}{2},$ find $2x^2+1$.

My attempt: $\cos (\sin^{-1}(x)+\tan^{-1}(x)) = \sqrt{1-x^2}\sqrt{\frac{1}{1+x^2}}-\frac{x^2}{\sqrt{x^2+1}} = 0.$

So we have $1-x^2=x^2$ which means $x= \frac{1}{\sqrt{2}}$which would mean $2x^2+1=2$.However, my book says that the answer is $\sqrt{5}$.

Where am I messing up?

Source:- BITSAT 2019

Answer 1

We don't have $1-x^2=x^2$ but $\sqrt{1-x^2}=x^2$ which means $x^2=\frac{-1+\sqrt5}2$ hence $2x^2+1=\sqrt5.$

Answer 2

Given $$ \sin^{-1} x + \tan ^{-1} x = \pi/2$$

Trig identity $$ \sin^{-1} x + \cos ^{-1} x = \pi/2$$

Subtract $$ \tan^{-1} x = \cos ^{-1} x $$

Take cos both sides and simplify using a right triangle construction

$$ x = \cos ( \tan ^{-1} x )= \frac{1} {\sqrt{1+x^2}}$$

Square and simplify

$$ x^4+x^2 -1=0,\text{ Solve quadratic },~ x^2=\frac{-1\pm \sqrt 5}{2} $$

$$ 2 x^2+1= \pm \sqrt 5 $$

So there is one more RHS with a negative sign but that may be implied.