Question

If ${\sin }^{-1}x+{\tan }^{-1}x=\frac{\pi }{2}$ then prove that: $2x^2=\sqrt{5}-1$

Answer 1

${\tan }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}x$
${\tan }^{-1}x={\cos }^{-1}x$
${\tan }^{-1}x={\tan }^{-1}\frac{\sqrt{1-x^2}}{x}$
$x=\frac{\sqrt{1-x^2}}{x}$
$x^2=\sqrt{1-x^2}$
On squaring both sides
$x^4=1-x^2$
$x^4-x^2-1=0$
$x^2=\frac{-1\pm \sqrt{1+4}}{2}$
There are two possibilities solution -ve and +ve
When -ve taken $x^2=\frac{-1-\sqrt{5}}{2}$
When +ve taken $x^2=\frac{-1+\sqrt{5}}{2}$
$2x^2=\sqrt{5}-1$
Hence proved