Question

If $y={\cos }^{-1}\frac{a+b\cos x}{b+a\cos x},b>a,$ then$\frac{dy}{dx}$ at $x=0,$ is

• A. $\frac{\sqrt{b^2-a^2}}{b+a}$
• B. $\sqrt{\frac{b-a}{b+a}}$
• C. $\sqrt{\frac{b+a}{b-a}}$
• D. $\frac{\sqrt{b^2+a^2}}{b+a}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{\sqrt{b^2-a^2}}{b+a}$
B $\sqrt{\frac{b-a}{b+a}}$

$y={\cos }^{-1}\frac{a+b\cos x}{b+a\cos x},b>a,$
$\cos y=\frac{a+b\cos x}{b+a\cos x}$
$\therefore \log \cos y=\log (a+b\cos x)-\log (b+a\cos x)$
$\therefore \frac{1}{\cos y}(-\sin y)\frac{dy}{dx}=\frac{-b\sin x}{a+b\cos x}+\frac{a\sin x}{b+a\cos x}$ (i)
Now at $x=0,y={\cos }^{-1}1=0$ Clearly at point $(0,0)$ L.H.S of (i) is zero.