Question

The value of ${\left(\cot \frac{x}{2}-\tan \frac{x}{2}\right)}^2\left(1-2\tan x\cot 2x\right)$ is
(a) 1
(b) 2
(c) 3
(d) 4

Answer 1

(d) 4

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {\left(\cot \frac{x}{2}-\tan \frac{x}{2}\right)}^2\left(1-2\tan x\cot 2x\right) \\ \left({\cot }^2\frac{x}{2}-2\cot \frac{x}{2}\tan \frac{x}{2}+{\tan }^2\frac{x}{2}\right)\left\{1-2\tan x\left(\frac{{\cot }^{2}x-1}{2\cot x}\right)\right\} \\ \left({\cot }^2\frac{x}{2}-2+{\tan }^2\frac{x}{2}\right)\left\{1-\tan x\left(\frac{{\cot }^{2}x-1}{\cot x}\right)\right\} \\ \left({\cot }^2\frac{x}{2}+{\tan }^2\frac{x}{2}-2\right)\left(1-\frac{\cot x-\tan x}{\cot x}\right) \\ \left({\cot }^2\frac{x}{2}+{\tan }^2\frac{x}{2}-2\right)\left({\tan }^{2}x\right) \\ \left({\cot }^2\frac{x}{2}+{\tan }^2\frac{x}{2}-2\right){\left(\frac{2\tan {\displaystyle \frac{x}{2}}}{1-{\tan }^2{\displaystyle \frac{x}{2}}}\right)}^2 \end{gathered}$$
$$\begin{gathered} =\frac{1}{{\left(1-{\tan }^2{\displaystyle \frac{x}{2}}\right)}^2}\left(4+4{\tan }^4\frac{x}{2}-8{\tan }^2\frac{x}{2}\right) \\ =\frac{1}{{\left(1-{\tan }^2{\displaystyle \frac{x}{2}}\right)}^2}\left(4-8{\tan }^2\frac{x}{2}+4{\tan }^4\frac{x}{2}\right) \\ =\frac{4}{{\left(1-{\tan }^2{\displaystyle \frac{x}{2}}\right)}^2}\left\{{\left({\tan }^2\frac{x}{2}\right)}^2-2\left({\tan }^2\frac{x}{2}\right)+1\right\} \\ =\frac{4{\left({\tan }^2{\displaystyle \frac{x}{2}}-1\right)}^2}{{\left(1-{\tan }^2{\displaystyle \frac{x}{2}}\right)}^2} \\ =4 \end{gathered}$$