Let x = cos 2θ ⇒ θ =

1

2

cos−1 x ... 1
∴ √1+x = √1+cos2θ = √1+2cos2θ−1 = √2 cos θ
√1−x = √1−cos2θ = √1−1−2sin2θ = √2 sin θ
Let y = tan−1|

√1+x−√1−x

√1+x+√1−x

|
= tan−1|

√2cosθ−√2sinθ

√2cosθ+√2sinθ

|
= tan−1|

1−tanθ

1+tanθ

|
= tan−1{tan(

π

4

−θ)}
=

π

4

- θ =

π

4

-

1

2

cos−1 x From 1
∴

dy

dx

= −

1

2

(−

1

√1−x2

) =

1

2√1−x2