Prove sin theta-2sin cube theta/2cos cube theta-cos theta=tan theta
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Answered by
1831
(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
=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
Answered by
706
Answer:
Step-by-step explanation:
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.
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