Question

tan2A-tanA =
2sinA/cosA+cos3A​

Answer 1

tan2A-tanA =  2sinA/(cosA+cos3A)​

Step-by-step explanation:

tan2A-tanA =  2sinA/(cosA+cos3A)​

LHS = tan2A - tanA

= 2tanA /(1- tan²A)  - tanA

= tanA ( 2/(1 - tan²a)  - 1 )

= tanA( (1 +  tan²A)/(1 - tan²A))

= tanA ( 1 + sin²A/Cos²A)/( 1 - Sin²A/Cos²A))

= tanA ( cos²A + Sin²A)/(cos²A - Sin²A)

=tanA ( 1 / (cos²A - Sin²A)

= tanA/Cos2A

RHS =  2Sina/(cosA + Cos3A)

= 2SinA / ( CosA  + 4Cos³A - 3CosA)

= 2SinA / (4Cos³A - 2CosA)

= 2SinA/( 2CosA(2Cos²A - 1)

= tanA / Cos2A

LHS = RHS

QED

Proved

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Answer 2

L.H.S.=R.H.S.

Step-by-step explanation:

$\tan 2A-\tan A = \frac{2\sin A}{\cos A+\cos 3A}$

$L.H.S. = \tan 2A-\tan A$        $\left \{ \because \tan 2A =\frac{2\tan A}{1-\tan ^{2}A} \right \}$

$= \frac{2\tan A}{1-\tan ^{2}A}-\tan A$

$=\tan A\left \{ \frac{(2-1+\tan ^{2}A)}{1-\tan ^{2}A} \right \}$

$= \frac{\tan A}{\cos 2A}$

$= \frac{\sin A}{\cos A \cos 2A}$

$=\frac{2\sin A}{2\cos 2A \cos A}$ $\left \{ \because 2\cos C\cos D = \cos(C+D)+\cos(C-D) \right \}$

$=\frac{2\sin A}{\cos A+\cos 3A}$

Hence, L.H.S.=R.H.S.