Question

8. sec2 2x1- tan 2x

Answer 1

Given that,

sec 2 2x=1−tan2x 1+ tan 2 2x=1−tan2x            ( ∵ sec 2 x=1+ tan 2 x ) tan 2 2x+tan2x=0 tan2x( tan2x+1 )=0

Hence,

tan2x=0

tan2x+1=0 tan2x=−1

General solution for tan2x=0 ,

Consider,

tanx=tany tan2x=tan2y (1)

Also,

tan2x=0 (2)

From equations (1) and (2),

tan2y=0 y=0

General solution is given by,

2x=nπ+2y 2x=nπ+0 x= nπ 2

General solution for tan2x=−1 Consider,

tan2x=tan2y (1)

Here,

tan2x=−1 (2)

From equations (1) and (2),

tan2y=−1 tan( 2y )=tan( 3π 4 ) 2y= 3π 4

General solution is given by,

2x=nπ+2y  ( n∈z )

For, 2y= 3π 4

2x=nπ+ 3 4 π x= nπ 2 + 3 8 π

Where, n∈z

Thus, the general solutions are,

For tan2x=0, x= nπ 2

For tan2x=−1, x= nπ 2 + 3π 8 where, n∈z