Question

Prove using identities:-

$\dfrac{1 + {tan}^{2} \theta }{1 + {cot}^{2} \theta} ={ ( \dfrac{1 + tan \theta}{1 + cot \theta} ) }^{2} = {tan}^{2} \theta \\ \\ \ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ \begin{gathered} \tt \red{!!!! \: \: No \: spam \: \: !!!! } \\ \sf{\underline{⋆ \: Proper \: explanation \: ⋆}}\end{gathered}$

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Answer 1

Hope it's helpful for you ✌

Answer 2

1+cot2θ1+tan2θ=1+tan2θ11+tan2θ

(1−cotθ1−tanθ)2=(tan2θtan2θ+1)1+tan2θ

=⎝⎜⎜⎛1−tanθ11−tanθ⎠⎟⎟⎞2=tan2θ

(tanθ(tanθ−11−tanθ))2

=(−tanθ)2

=tan2θ.

hope it help uh dude