Question

hola!!

answer it with clear explanation....​

Answer 1

Refer to the attached file :)

Answer 2

$\large{\underline{\underline{\mathfrak{\pink{\sf{Prove\:here-}}}}}}$.

✴$\bold{\frac{p^2-1}{p^2+1}\:=\:sin\theta}$.

$\large{\underline{\underline{\mathfrak{\green{\sf{Given\:here-}}}}}}$.

✴$\:Sec\theta \:+\:tan\theta\:=\:p$..................(1).

➡First find here.

$\implies\bold{\red{\frac{p^2-1}{p^2+1}}}$.

➡Keep values by (1).

$\implies\fr{\frac{(\:Sec\theta\:+\:tan\theta)^2\:-\:1}{(\:Sec\theta\:+\:tan\theta)^2\:+\:1}}$.

$\implies\fr{\frac{(\:Sec\theta\:+\:tan\theta)^2\:-\:(\:sec^2\theta\:-\:tan^2\theta)}{(\:Sec\theta\:+\:tan\theta)^2\:+\:(\:Sec^2\theta\:-\:tan^2\:theta)}}$.

$\implies\fr{\frac{(\:sec\theta\:+\:tan\theta)\:(\:Sec\theta\:+\:tan\theta)\:-\:(\:sec\theta\:-\:tan\theta)}{(\:sec\theta\:+\:tan\theta)\:(\:Sec\theta\:+\:tan\theta)\:+\:(\:sec\theta\:-\:tan\theta)}}$.

$\implies\fr{\frac{(\:sec\theta\:+\:tan\theta)\:-\:(\:Sec\theta\:-\:tan\theta)}{(\:Sec\theta\:+\:tan\theta)\:+\:(\:Sec\theta\:-\:tan\theta)}}$.

$\implies\fr{\frac{\:Sec\theta\:+\:tan\theta\:-\:Sec\theta\:+\:tan\theta}{\:sec\theta\:+\:tan\theta\:+\:Sec\theta\:-\:tan\theta)}}]$.

$\implies\fr{\frac{\:2tan\theta}{\:2sec\theta}}$.

$\implies\fr{\frac{\:tan\theta}{\:sec\theta}}$.

$\implies\fr{\frac{\frac{\:sin\theta}{\:cos\theta}}{\frac{1}{\:Cos\theta}}}$.

$\implies\fr{\frac{\:sin\theta}{\:Cos\theta}\:*\:Cos\theta}$.

$\implies\bold{\:Sin\theta}$.

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