Question

solve fast!.
no spam! answer if u kmkw​

Answer 1

see this,

hope this will help you !!

thanks!

Answer 2

Tanθ+sinθ=m

tanθ-sinθ=n

∴, m+n=tanθ+sinθ+tanθ-sinθ=2tanθ

m-n=tanθ+sinθ-tanθ+sinθ=2sinθ

mn=(tanθ+sinθ)(tanθ-sinθ)

     =tan²θ-sin²θ

∴, m²-n²

=(m+n)(m-n)

=2tanθ.2sinθ

=4sinθtanθ

4√mn

=4√(tan²θ-sin²θ)

=4√(sin²θ/cos²θ-sin²θ)

=4√sin²θ(1/cos²θ-1)

=4sinθ√(1-cos²θ)/cos²θ

=4sinθ/cosθ√sin²θ [∵, sin²θ+cos²θ=1]

=4sinθtanθ

∴, LHS=RHS (proved)

Hope it helps...

please make brainliest...!