Question

How to prove the above using trigonometric identities​

Answer 1

Answer:

Step-by-step explanation:

to prove - $\frac{sin A - 2sin^{3} A }{2cos^{3} A - cosA } = tanA$

LHS = $\frac{sin A - 2sin^{3} A }{2cos^{3} A - cosA }$

divide the all ratios by cosA we get,

                       $\frac{tanA - 2tanAsin^{2} A }{2cos^{2} A - 1 }$

                  =  $\frac{tanA(1-2sin^{2} A)}{2(1-sin^{2} A)-1}$

                  = $\frac{tanA(1-2sin^{2} A)}{1-2sin^{2} A}$

                  = $tanA$ hence prove

if you don't understand ask me again i will explain this in detail ok