Question

prove that sin 3 theta - sin theta ÷ cos theta - cos 3 theta = cot 2 theta

Answer 1

proof.................

Answer 2

Answer with Step-by-step explanation:

To prove: $\dfrac{sin3\theta -sin\theta}{cos\theta -cos3\theta} = cot2\theta$

As we know:

$sinx-siny =2cos\dfrac{x+y}{2} sin \dfrac{x-y}{2} \\\\\\cosx-cosy=-2sin\dfrac{x+y}{2} sin\dfrac{x-y}{2}$

$L.H.S= \dfrac{sin3\theta-sin\theta}{cos\theta-cos3\theta} =\dfrac{2cos\dfrac{3\theta+\theta}{2} sin \dfrac{3\theta-\theta}{2}}{-2sin\dfrac{\theta+3\theta}{2} sin\dfrac{\theta-3\theta}{2}} \\\\\\\\=\dfrac{cos2\theta sin2\theta}{-sin2\theta(-sin2\theta) } =\dfrac{cos2\theta}{sin2\theta} =cot2\theta=R.H.S$

Hence proved the required result