Question : The value of $\left(2 \cos ^2 \theta-1\right)\left[\frac{1+\tan \theta}{1-\tan \theta}+\frac{1-\tan \theta}{1+\tan \theta}\right]$ is:
Option 1: $2$
Option 2: $0$
Option 3: $\frac{\sqrt{3}}{2}$
Option 4: $1$
Correct Answer: $2$
Solution :
Given expression,
$\left(2 \cos ^2 \theta-1\right)\left[\frac{1+\tan \theta}{1-\tan \theta}+\frac{1-\tan \theta}{1+\tan \theta}\right]$
We know, $\tan\theta=\frac{\sin\theta}{\cos\theta}$
= $(2\cos^2\theta-(\sin^2\theta+\cos^2\theta))\left[\frac{1+\frac{\sin\theta}{\cos\theta}}{1-\frac{\sin\theta}{\cos\theta}}+\frac{1-\frac{\sin\theta}{\cos\theta}}{1+\frac{\sin\theta}{\cos\theta}}\right]$
= $(2\cos^2\theta-\sin^2\theta-\cos^2\theta))\left[\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}+\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\right]$
= $(\cos^2\theta-\sin^2\theta)\left[\frac{(\cos\theta+\sin\theta)^2+(\cos\theta-\sin\theta)^2}{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}\right]$
= $(\cos^2\theta-\sin^2\theta)\left[\frac{\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta+\sin^2\theta-2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}\right]$
= $1+1$ [As $\cos^2\theta+\sin^2\theta=1$]
= $2$
Hence, the correct answer is $2$.
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