Question

If $0<{\cos }^{-1}x<1$ and $1+\sin \left({\cos }^{-1}x\right)+{\sin }^2\left({\cos }^{-1}x\right)+{\sin }^3\left({\cos }^{-1}x\right)+...infinity=2$ then the value of $12x^2$ is

Answer 1

The above equation is an infinite G.P
Hence sum of the G.P will be

$S$ $=\frac{a}{1-r}$

$=\frac{1}{1-\sin \left({\cos }^{-1}\right(x\left)\right)}$ ...(i)

Let ${\cos }^{-1}\left(x\right)=y$
$\therefore \cos y=x$
$\sin y=\pm \sqrt{1-x^2}$

$y={\sin }^{-1}(\pm \sqrt{1-x^2})$

$=\pm {\sin }^{-1}\left(\sqrt{1-x^2}\right)$
Substituting in the equation, gives us

$=\frac{1}{1\pm \sqrt{1-x^2}}$
$=2$
$2\pm 2\sqrt{1-x^2}=1$
$2-1=\pm 2\sqrt{1-x^2}$

$\frac{1}{2}=\pm \sqrt{1-x^2}$

Squaring both sides, we get

$\frac{1}{4}=1-x^2$

$x^2=\frac{3}{4}$

Hence, $12x^2$ $=12\times \frac{3}{4}$ $=9$.