Let f (x) =

secx−1

secx+1

⇒ f (x) =

1 cosx −1

1 cosx +1

⇒ f (x) =

1−cosx

1+cosx

... (i)
Differentiating (i) with respect to x, we get

d

dx

(f (x)) =
=
=
=

2sinx

(1+cosx)2

=

2 1 cosecx

(1+ 1 secx )2

=

2 cosecx

(secx+1)2 sec2x

=

2

cosecx

.

sec2x

(sec+1)2

=

2sinx. 1 cosx .secx

(secx+1)2

=

2tanx.secx

(secx+1)2