Question

If $0<{\cos }^{-1}x<1$ and $1+{\cos }^{-1}x+({\cos }^{-1}x{)}^2+\dots \dots \infty =2,$ then $x$ is equal to

• A. $\frac{\pi }{4}$
• B. $\cos \frac{1}{2}$
• C. $\cos \frac{1}{\sqrt{2}}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\cos \frac{1}{2}$

$1+{\cos }^{-1}x+({\cos }^{-1}x{)}^2+\dots \dots \infty =2$
This is an infinite geometric progression
$\Rightarrow \frac{1}{1-{\cos }^{-1}x}=2\Rightarrow {\cos }^{-1}x=\frac{1}{2}\therefore x=\cos \frac{1}{2}$