Prove the following trigonometric identities:$ \frac{\sin A-2 \sin ^{3} A}{2 \cos ^{3} A-\cos A}=\tan A $

To do:

We have to prove that \( \frac{\sin A-2 \sin ^{3} A}{2 \cos ^{3} A-\cos A}=\tan A \).

Solution:

We know that,

$\tan A=\frac{\sin A}{\cos A}$.....(i)

$\cos ^{2} A+\sin^2 A=1$.......(ii)

Therefore,

$\frac{\sin A-2 \sin ^{3} A}{2 \cos ^{3} A-\cos A}=\frac{\sin A(1-sin^2 A)}{\cos A(2cos^2 A-1)}$               

$=\tan A(\frac{1-2sin^2 A}{2cos^2 A-1})$                       [From (i)]

$=\tan A(\frac{\cos^2 A+\sin^2 A-2\sin^2 A}{2\cos^2 A-\cos^2 A-\sin^2 A})$             [From (ii)]

$=\tan A(\frac{\cos^2 A-\sin^2 A}{\cos^2 A-\sin^2 A})$                   

$=\tan A$              

Hence proved.    

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Updated on: 10-Oct-2022

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