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Question
If 2 tan−1 (cos θ) = tan−1 (2 cosec θ), (θ ≠ 0), then find the value of θ.
Solution
\[2 \tan^{- 1} \left( \cos\theta \right) = \tan^{- 1} \left( 2cosec\theta \right)\]
\[ \Rightarrow \tan^{- 1} \left( \frac{2\cos\theta}{1 - \cos^2 \theta} \right) = \tan^{- 1} \left( 2cosec\theta \right) \left[ 2 \tan^{- 1} x = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) \right]\]
\[ \Rightarrow \frac{2\cos\theta}{1 - \cos^2 \theta} = 2cosec\theta\]
\[ \Rightarrow \frac{\cos\theta}{1 - \cos^2 \theta} = \frac{1}{\sin\theta}\]
\[ \Rightarrow 1 - \cos^2 \theta = \sin\theta\cos\theta\]
\[ \Rightarrow \sec^2 \theta - 1 = \tan\theta \left[ \text { Dividing both sides by } \cos^2 \theta \right]\]
\[ \Rightarrow 1 + \tan^2 \theta - 1 = \tan\theta\]
\[ \Rightarrow \tan^2 \theta - \tan\theta = 0\]
\[ \Rightarrow \tan\theta\left( \tan\theta - 1 \right) = 0\]
\[ \Rightarrow \tan\theta = 0 or \tan\theta - 1 = 0\]
\[ \Rightarrow \tan\theta = 0 or \tan\theta = 1\]
\[ \Rightarrow \theta = 0 or \theta = \frac{\pi}{4}\]
It is given that θ ≠ 0
\[\therefore \theta = \frac{\pi}{4}\]
Thus, the value of θ is \[\frac{\pi}{4}\] .
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