Value of \((\cot\frac{x}{2}-\tan\frac{x}{2})^2(1-2\tan x\cot 2x)\)
Question
Find the value of
\[ \left(\cot\frac{x}{2}-\tan\frac{x}{2}\right)^2 \left(1-2\tan x\cot 2x\right) \]
(a) 1
(b) 2
(c) 3
(d) 4
Solution
Using the identity
\[ \cot\theta-\tan\theta = \frac{\cos^2\theta-\sin^2\theta} {\sin\theta\cos\theta} = \frac{\cos2\theta}{\frac12\sin2\theta} = 2\cot2\theta \]
Putting \(\theta=\frac{x}{2}\),
\[ \cot\frac{x}{2}-\tan\frac{x}{2} = 2\cot x \]
Therefore,
\[ \left(\cot\frac{x}{2}-\tan\frac{x}{2}\right)^2 = 4\cot^2x \]
Now,
\[ 1-2\tan x\cot2x \]
Using
\[ \cot2x = \frac{\cos2x}{\sin2x} = \frac{\cos2x}{2\sin x\cos x} \]
\[ 2\tan x\cot2x = 2\cdot\frac{\sin x}{\cos x} \cdot \frac{\cos2x}{2\sin x\cos x} = \frac{\cos2x}{\cos^2x} \]
Hence,
\[ 1-2\tan x\cot2x = 1-\frac{\cos2x}{\cos^2x} \]
Using \(\cos2x=2\cos^2x-1\),
\[ = 1-\frac{2\cos^2x-1}{\cos^2x} = \frac{1-\cos^2x}{\cos^2x} = \tan^2x \]
Therefore,
\[ 4\cot^2x\cdot\tan^2x = 4 \]
Final Answer
\[ \boxed{4} \]
Hence, the correct option is (d) 4.