Question

Question 7) Sin theta - 2 sin cube theta divided by 2 cos cube theta - cos theta= tan theta...prove it

Answer 1

sinθ-2sin³θ/2cos³θ-cosθ

Taking common sin θ from numerator

Taking common cos θ from Denominator

sinθ(1-2sin²θ)/cosθ(2cos²θ-1)

sinθ/cosθ=tanθ

sin²θ + cos²θ=1

SO, substitute 1 as sin²θ + cos²θ

tanθ(sin²θ + cos²θ-2sin²θ)/(2cos²θ-sin²θ -cos²)

tanθ( cos²θ-sin²θ)/( cos²θ-sin²θ)

tanθ=RHS

Hence Proved

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