Question

If $\mathrm{f}\left(\mathrm{x}\right)=\frac{(1-\mathrm{x})}{(1+\mathrm{x})}$then $\mathrm{f}\left(\mathrm{f}\right(\cos 2\mathrm{\theta }\left)\right)=$

• A. $\tan \left(2\mathrm{\theta }\right)$
• B. $\sec \left(2\mathrm{\theta }\right)$
• C. $\cos \left(2\mathrm{\theta }\right)$
• D. $\cot \left(2\mathrm{\theta }\right)$

Answer 1

The correct option is C

$\cos \left(2\mathrm{\theta }\right)$

Explanation for the correct otpion:

Step 1: solving f($\cos 2\mathrm{\theta }$)

Given: $\mathrm{f}\left(\mathrm{x}\right)=\frac{(1-\mathrm{x})}{(1+\mathrm{x})}$

$\mathrm{f}\left(\cos 2\mathrm{\theta }\right)=\frac{1-\cos 2\mathrm{\theta }}{1+\cos 2\mathrm{\theta }}$

We know that,

$\begin{array}{rcl}\therefore \cos 2\mathrm{\theta }& =& 2{\cos }^2\mathrm{\theta }-1\\ & =& 1-2{\sin }^2\mathrm{\theta }\end{array}$

Substitute in the above equation,

$\begin{array}{rcl}\therefore \mathrm{f}\left(\cos 2\mathrm{\theta }\right)& =& \frac{1-(1-2{\sin }^2\mathrm{\theta })}{1+(2{\cos }^2\mathrm{\theta }-1)}\\ & =& \frac{2{\sin }^2\mathrm{\theta }}{2{\cos }^2\mathrm{\theta }}\\ & =& \frac{{\sin }^2\mathrm{\theta }}{{\cos }^2\mathrm{\theta }}\\ & =& {\tan }^2\mathrm{\theta }\end{array}$

Step 2: solving f(f($\cos 2\mathrm{\theta }$))

$\begin{array}{rcl}\mathrm{f}\left(\mathrm{f}\right(\cos 2\mathrm{\theta })& =& \frac{1-\mathrm{f}\left(\cos 2\mathrm{\theta }\right)}{1+\mathrm{f}\left(\cos 2\mathrm{\theta }\right)}\\ & =& \frac{1-{\tan }^2\mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}\\ & =& \frac{1-{\displaystyle \frac{{\sin }^2\mathrm{\theta }}{{\cos }^2\mathrm{\theta }}}}{1+{\displaystyle \frac{{\sin }^2\mathrm{\theta }}{{\cos }^2\mathrm{\theta }}}}\\ & =& \frac{{\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }}{{\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }}\\ & =& \frac{\cos 2\mathrm{\theta }}{1}\left[\because {\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }=\cos 2\mathrm{\theta },{\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }=1\right]\\ \mathrm{f}\left(\mathrm{f}\right(\cos 2\mathrm{\theta })& =& \cos 2\mathrm{\theta }\\ & & \end{array}$

Hence, the correct option is an option(c),