Question

Question number (7). It’s a prooving question
Plzzz answer fast ...it’s an emergency...I will mark it the brainliest answer..plz guyz

Answer 1

Here is ur answer......

Answer 2

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