Question

If ${\cos }^{-1}x=a+{\tan }^{-1}\frac{\sqrt{1-x^b}}{cx}$ for $x<0$.
Find the value of a ,b and c.

• A. $a=\pi ,b=1,c=1$
• B. $a=-\pi ,b=2,c=2$
• C. $a=-\pi ,b=1,c=2$
• D. $a=\pi ,b=2,c=1$

Answer 1

The correct option is D $a=\pi ,b=2,c=1$

Let $x=\cos \theta$

$\Rightarrow \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\sqrt{1-x^2}}{x}$ (i)

$\therefore {\cos }^{-1}x=\pi +\theta$ since $x<0$

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

Comparing with

${\cos }^{-1}x=a+{\tan }^{-1}\frac{\sqrt{1-x^b}}{cx}$ for $x<0$

$a=\pi ,b=2,c=1$