Question

${\cos }^{-1}\left(\frac{3+5\cos x}{5+3\cos x}\right)$ is equal to:

• A. ${\tan }^{-1}\left(\frac{1}{2}\tan \frac{x}{2}\right)$
• B. $2{\tan }^{-1}\left(2\tan \frac{x}{2}\right)$
• C. $\frac{1}{2}{\tan }^{-1}\left(2\tan \frac{x}{2}\right)$
• D. $2{\tan }^{-1}\left(\frac{1}{2}\tan \frac{x}{2}\right)$
• E. ${\tan }^{-1}\left(\tan \frac{x}{2}\right)$

Answer 1

Answer: (D)

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