Question

(vii) (31) tan 3A = 3 tan A - tan³ A
/1 - 3 tan² A​

Answer 1

To prove --->

tan3A = (3tanA - tan³A ) / ( 1 - 3tan²A )

Proof---> We have a formula

tan(A + B ) = (tanA + tanB ) / ( 1 - tanA tanB )

LHS = tan 3A

We can break 3A as ( 2A + A )

= tan ( 2A + A )

Now applying above formula we get

= ( tanA + tan2A ) / ( 1 - tanA tan2A )

We have another formula as follows

tan2A = 2tanA / ( 1 - tan²A ) , applying it here we get.

tanA + ( 2tanA / 1 - tan²A )

= ---------------------------------------

1 - tanA ( 2tanA / 1 - tan²A )

Taking ( 1 - tan²A ) as LCM , we get

{ tanA ( 1 - tan²A ) + 2tanA } / ( 1 - tan²A )

= ---------------------------------------------------------

{ 1 - tan²A - 2tan²A } / ( 1 - tan²A )

tanA - tan³A + 2 tanA

= ---------------------------------

1 - 3 tan²A

= ( 3 tanA - tan³A ) / ( 1 - 3 tan²A ) = RHS

Additional information--->

(1) Sin2A = 2SinA CosA

(2) Cos2A = 2Cos²A - 1

= 1 - 2 Sin²A

= Cos²A - Sin²A

(3) Sin2A = 2tanA / (1 + tan²A )

(4 ) Cos2A = ( 1 + tan²A ) / ( 1 - tan²A )

(5 ) Sin3A = 3SinA - 4Sin³A

( 6 ) Cos3A = 4Cos³A - 3CosA