6 पष्य » _. 8 809 - 6 S00Z ६७_— 0 UIS z - g uls

1.	6 पष्य » _. 8 809 - 6 S00Z ६७_— 0 UIS z - g uls
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 please like the solution 👍 ✔️

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

please like the solution 👍 ✔️

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