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Question Number 211956 by Durganand last updated on 25/Sep/24

Answered by Frix last updated on 25/Sep/24

$$\tan \alpha =t\tan 2\alpha =\frac{2t}{1-t^2}\sin 2\alpha =\frac{2t}{1+t^2}$$$$\frac{1}{t}-\frac{1-t^2}{2t}=\frac{2-(1-t^2)}{2t}=\frac{1+t^2}{2t}=\frac{1}{\sin 2\alpha }$$

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