Question

plzzzzzxzz solve.......

Answer 1

Hey,


(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

HOPE IT HELPS YOU:-))