1 + (1/tan^2 x) = 1/(sin^2 x)

We will use trigonometric identities to simplify.

We know that tanx = sinx/cosx

==> 1+ (1/tan^2 x ) = 1+ (1/(sin^2x/cos^2 x)

= 1+ (cos^2x/sin^2x)

Now we will rewrite sin^2 x/sin^2 x = 1

==> 1+ (1/tan^2 x)= sin^2 x/sin^2 x + cos^2 x/sin^2 x

= (sin^2 x + cos^2 x)/sin^2 x

= 1/sin^2 x..........q.e.d

**==> Then the identity is true.**

We have to prove that 1 + (1/(tan x)^2) = 1/(sin x)^2.

Use the relation that tan x = sin x / cos x.

1 + (1/(tan x)^2)

=> 1 + (1/((sin x)^2/(cos x)^2))

=> 1 + (cos x)^2/(sin x)^2

=> [(sin x)^2 + (cos x)^2]/(sin x)^2

=> 1/(sin x)^2

This is true for all values of x except x = k*pi as the value of sin x and tan x is 0 and 1/0 is not defined.

**We prove 1 + (1/(tan x)^2) = 1/(sin x)^2**