Question

4tan x (1- tan21-6 tan2x tan4x23.tan 4x =x)

Answer 1

The given trigonometric function is tan4x= 4tanx( 1− tan 2 x ) 1−6 tan 2 x+ tan 4 x .

The formula of trigonometric identity cosA−cosB is given by,

tan2A= 2tanA 1− tan 2 A

Solve the left hand side of the trigonometric function.

L.H.S.=tan4x =tan2( 2x ) = 2tan( 2x ) 1− tan 2 ( 2x ) = 2( 2tanx 1− tan 2 x ) 1− ( 2tanx 1− tan 2 x ) 2

Further simplify the above expression.

L.H.S.= ( 4tanx 1− tan 2 x ) [ 1− 4 tan 2 x ( 1− tan 2 x ) 2 ] = ( 4tanx 1− tan 2 x ) [ ( 1− tan 2 x ) 2 −4 tan 2 x ( 1− tan 2 x ) 2 ] = 4tanx( 1− tan 2 x ) ( 1− tan 2 x ) 2 −4 tan 2 x = 4tanx( 1− tan 2 x ) 1+ tan 4 x−2 tan 2 x−4 tan 2 x

Further simplify the above expression.

L.H.S.= 4tanx( 1− tan 2 x ) 1+ tan 4 x−6 tan 2 x = 4tanx( 1− tan 2 x ) 1−6 tan 2 x+ tan 4 x =R.H.S.

This is equal to the right hand side of the given trigonometric expression.

Thus, the trigonometric function is tan4x= 4tanx( 1− tan 2 x ) 1−6 tan 2 x+ tan 4 x .

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

Hence proved.