Question

11. tanCOS X1-x42V2

Answer 1

We have to prove that tan −1 ( 1+x − 1−x 1+x + 1−x )= π 4 − 1 2 cos −1 x, − 1 2 ≤x≤1.

Consider that x=cos2θ, then, θ= 1 2 cos −1 x.

Substitute x=cos2θ to the left hand side of the given equation.

tan −1 ( 1+x − 1−x 1+x + 1−x )= tan −1 ( ( 1+cos2θ − 1−cos2θ ) ( 1+cos2θ )+( 1−cos2θ ) ) = tan −1 ( 2 cos 2 θ − 2 sin 2 θ 2 cos 2 θ + 2 sin 2 θ ) = tan −1 ( 2 cosθ− 2 sinθ 2 cosθ+ 2 sinθ ) = tan −1 ( cosθ−sinθ cosθ+sinθ )

Further solving,

tan −1 ( 1+x − 1−x 1+x + 1−x )= tan −1 ( 1−tanθ 1+tanθ ) = tan −1 1− tan −1 ( tanθ ) = π 4 −θ = π 4 − 1 2 cos −1 x

Hence, it is proved that tan −1 ( 1+x + 1−x 1+x − 1−x )= π 4 − 1 2 cos −1 x.