Question

plz give the step by step solution​

Answer 1

Answer:-

We have to prove:

$\sf \: \frac{ \sin\theta - 2 { \: \sin }^{3} \theta}{2 \: { \cos }^{3} \theta - \cos\theta} = \tan \theta$

Taking Sin ∅ / Cos ∅ in LHS we get,

$\sf \implies \: \frac{ \sin\theta(1 - 2 { \sin }^{2} \theta)}{ \cos \theta( { 2\cos }^{2} \theta - 1)} = \tan \theta \\ \\ \implies \: \tan \theta(\frac{1 - 2 { \sin}^{2} \theta}{ { 2\cos}^{2} \theta - 1 } )= \tan \theta$

We know that,

Sin² ∅ + Cos² ∅ = 1

→ Sin² ∅ = 1 - Cos² ∅

Putting the value of Sin² ∅ we get,

$\sf \implies \: \tan \theta \: ( \frac{1 - 2(1 - { \cos }^{2} \theta) }{2 { \cos}^{2} \theta - 1} ) = \tan\theta \\ \\ \sf \implies \: \tan \theta( \frac{1 - 2 + 2 { \cos }^{2} \theta)}{2 { \cos }^{2} \theta - 1} ) = \tan \theta \\ \\ \sf \implies \: \tan \theta( \frac{2 { \cos}^{2} \theta - 1 }{2 { \cos }^{2} \theta - 1 } ) = \tan \theta \\ \\ \sf \implies \: \tan \theta(1) = \tan \theta \\ \\ \sf \implies \: \tan \theta = \tan \theta \\ \\ \implies\sf \large{LHS = RHS.}$

Hence, Proved.