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

If \tan p θ = p \tan θ then prove that

\frac{\sin^{2} p θ}{\sin^{2} θ} = \frac{p^{2}}{1 + (p^{2} - 1) \sin^{2} θ} .

Answered by Frix last updated on 13/Apr/24

\tan p θ = p \tan θ = t

\Leftrightarrow

p θ = \tan^{- 1} t \land θ = \tan^{- 1} \frac{t}{p}

\Rightarrow

\sin^{2} p θ = \frac{t^{2}}{t^{2} + 1} \land \sin^{2} θ = \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}}} which is true

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