We have to prove : (1/(sin x^2x)+(1/cos^2x)=(tan x+(1/tan x))^2

Take the left hand side

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

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

=> 1/(sin x)^2*(cos x)^2x ...(1)

Take the right hand side

(tanx+(1/tanx))^2

use tan x = sin x/cos x

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

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

=> 1/(cos x)^2*(sin x)^2 ...(2)

From (1) and (2), we see that both the sides are the same.

**This proves that: (1/(sin x^2x)+(1/cos^2x)=(tan x+(1/tan x))^2**

We'll manage the right side by raising to square the binomial:

[tanx+(1/tanx)]^2 = (tan x)^2 + 2*tan x*(1/tan x) + 1/(tan x)^2

We'll simplify and we'll get:

[tanx+(1/tanx)]^2 = (tan x)^2 + 2 + 1/(tan x)^2

We'll write the middle terms as:

2 = 1 + 1

RHS = [tanx+(1/tanx)]^2 = [(tan x)^2 + 1] + [1 + 1/(tan x)^2]

RHS = 1/(cos x)^2 + [(tan x)^2 + 1]/(tan x)^2

RHS = 1/(cos x)^2 + 1/(cos x)^2*(tan x)^2

But (tan x)^2 = (sin x)^2/(cos x)^2

We'll multiply both sides by (cos x)^2:

(cos x)^2*(tan x)^2 = (sin x)^2

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

**We notice that RHS=LHS, therefore the identity 1/(cos x)^2 + 1/(sin x)^2 = [tanx+(1/tanx)]^2 is verified.**