Question

Which among the following represents simplified form of ${\cos }^{-1}\left(\frac{3+5\cos x}{5+3\cos x}\right)$?

• A. $2{\tan }^{-1}\left(\frac{1}{2}\tan \frac{x}{2}\right)$ when $x\in (0,\pi )$
• B. $-2{\tan }^{-1}\left(\frac{1}{2}\tan \frac{x}{2}\right)$ when $x\in (\pi ,2\pi )$
• C. ${\tan }^{-1}\left(\frac{1}{2}\tan \frac{x}{2}\right)$ when $x\in (2\pi ,3\pi )$
• D. $2{\tan }^{-1}\left(\tan \frac{x}{2}\right)$ when $x\in (3\pi ,4\pi )$

Answer 1

Answer: (A), (B)

$-2{\tan }^{-1}\left(\frac{1}{2}\tan \frac{x}{2}\right)$ when $x\in (\pi ,2\pi )$

Let $\theta ={\cos }^{-1}\frac{3+5\cos x}{5+3\cos x}$
$\Rightarrow \cos \theta =\frac{3+5\cos x}{5+3\cos x}\Rightarrow \frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=\frac{3+5\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}{5+3\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}\Rightarrow \frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=\frac{4-{\tan }^2\frac{x}{2}}{4+{\tan }^2\frac{x}{2}}\Rightarrow 4{\tan }^2\frac{\theta }{2}={\tan }^2\frac{x}{2}\Rightarrow \tan \frac{\theta }{2}=\left|\frac{1}{2}\tan \frac{x}{2}\right|\Rightarrow \theta =2{\tan }^{-1}\left|\frac{1}{2}\tan \frac{x}{2}\right|$

When $x\in (0,\pi )\cup (2\pi ,3\pi )$
$\theta =2{\tan }^{-1}\left(\frac{1}{2}\tan \frac{x}{2}\right)$
When $x\in (\pi ,2\pi )\cup (3\pi ,4\pi )$
$\theta =-2{\tan }^{-1}\left(\frac{1}{2}\tan \frac{x}{2}\right)$