I have to prove that: \(\displaystyle{{\tan}^{{2}}\theta}{{\sin}^{{2}}\theta}={{\tan}^{{2}}\theta}-{{\sin}^{{2}}\theta}\)

Samara Richard

Samara Richard

Answered question

2022-03-29

I have to prove that:
tan2θsin2θ=tan2θsin2θ

Answer & Explanation

umgebautv6v2

umgebautv6v2

Beginner2022-03-30Added 10 answers

tan2θsin2θ
=(sin2θcos2θ)(sin2θ)
=sin4θcos2θ
=sin2θsin2θcos2θ
=sin2θ(1cos2θ)cos2θ
=sin2θsin2θcos2θcos2θ
=sin2θcos2θsin2θcos2θcos2θ
=tan2θsin2θ
tan2θsin2θ=tan2θsin2θ

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?