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Question Number 206421 by MATHEMATICSAM last updated on 13/Apr/24

$$\mathrm{If}\tan p\theta =p\tan \theta \mathrm{then}\mathrm{prove}\mathrm{that}$$$$\frac{{\sin }^{2}p\theta }{{\sin }^2\theta }=\frac{p^2}{1+(p^2-1){\sin }^2\theta }.$$

Answered by Frix last updated on 13/Apr/24

$$\tan p\theta =p\tan \theta =t$$$$\iff$$$$p\theta ={\tan }^{-1}t\wedge \theta ={\tan }^{-1}\frac{t}{p}$$$$\Rightarrow$$$${\sin }^{2}p\theta =\frac{t^2}{t^2+1}\wedge {\sin }^2\theta =\frac{t^2}{t^2+p^2}$$$$\frac{\frac{t^2}{t^2+1}}{\frac{t^2}{t^2+p^2}}=\frac{p^2}{1+(p^2-1)\frac{t^2}{t^2+p^2}}\mathrm{which}\mathrm{is}\mathrm{true}$$$$$$

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