Question

proove sinQ-2sin^2Q@÷2cos^2Q-cosQ=tanQ​

Answer 1

Answer:

tanθ

Step-by-step explanation:

To prove--->There is a mistake in

-------------- question

Sinθ-2Sin³θ

---------------------- = tan θ

2Cos³θ-Cosθ

Solution--->

-------------

We know that, Sin²θ+Cos²θ=1

Shifting Cos²θ to RHS we get

=> Sin²θ=1-Cos²θ

Shifting Sin²θ to RHS we get

=> Cos²θ=1-Sin²θ

Now taking LHS of the given question

Sinθ - 2Sin³θ

L.H.S.=------------------------

2Cos³θ-Cosθ

Sinθ(1-2Sin²θ)

= ---------------------------

Cosθ(2Cos²θ-1)

Sinθ {1-2(1-Cos²θ)}

= ----------------------------------

Cosθ(2Cos²θ-1)

Sinθ (1-2+2Cos²θ)

=---------------------------------

Cosθ(2Cos²θ-1)

Sinθ (2Cos²θ-1)

= ------------------------------

Cosθ (2Cos²θ-1)

2Cos²θ-1 cancel out from numerator and denominator

Sinθ

= ---------

Cosθ

=tanθ =R.H.S.