Math, asked by udayuc, 10 months ago

plzzzz solveeeeeeeee​

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Answered by akhilesh220401
2

Answer:

sinθ- 2 sin^3θ)/(2cos^3θ-cosθ)

=sinθ(1-2sin^2θ)/cosθ(2cos^2 θ-1)

= sin θ [1- 2(1- cos^2 θ)/cos θ(2cos ^2 θ -1)

= sin θ [1-2+ 2cos ^2 θ]/ cos θ (2 cos^2θ -1)

= sin θ(2cos^2 θ -1)/ cos θ(2 cos ^2θ-1)

= sin θ/ cos θ

= tanθ

Read more on Brainly.in - https://brainly.in/question/878295#readmore

Answered by dezisantosh
1

Answer:

hope it helps

please mark as brainliest please dear....

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