Question

Sin A - 2 Sin cube A upon 2 cos cube A - Cos A is equal to tanA

Answer 1

I think it helps you

Answer 2

Hola there,

Given=>
$\frac{sin \: a - 2 {sin}^{3}a }{cos \: a - 2{cos}^{3}a } = tan \: a$

LHS
=> $\frac{sin \: a \: - 2{sin}^{3}a }{cos \: a \: - 2 {cos}^{3} a}$

=> $\frac{sin(1 - 2{sin}^{2} a)}{cos(1 - 2{cos}^{2} a)}$

=> $\frac{sin(1 - 2(1 - {cos}^{2} a)}{cos(1 - {cos}^{2} a)}$

=> $\frac{sin(1 - {cos}^{2} a)}{cos( 1 - {cos}^{2}a) }$

=> $\frac{sin \: a}{cos \: a}$

=> $tan \: a$

=> RHS

So, LHS = RHS

HENCE PROVED.

Hope this helps.....:)