Prove the identity: (1/sin^2x)+(1/cos^2x)=(tanx+(1/tanx))^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.