Question

Prove that
sin theta - 2 sin cube theta/ 2 cos cube theta - cos theta = tan theta​

Answer 1

Answer:

Hii mate :-)

LHS=(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 = RHS

HOPE IT HELPS,

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Answer 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

LHS = sinø(1-2sin^2ø)/cosø(2cos^2ø-1)

= sinø(1-2sin^2ø)/cos[2(1-sin^2ø)-1]

= sinø(1-2sin^2ø)/cosø(2-2sin^2ø-1)

= sinø(1-2sin^2ø)/cosø(1-2cosø)

= sinø/cosø = tanø = RHS

Hence proved.