# How can the following identity be proved : (tan x cot x)^2 = sec^2 x + csc^2 x

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### 2 Answers

We have to prove that (tan x * cot x)^2 = (sec x)^2 + (csc x)^2.

Let's start with the righ hand side:

(sec x)^2 + (csc x)^2

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

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

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

The left hand side is (tan x * cot x)^2

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

=> 1

It is not possible to equate (sin x)^2 * (cos x)^2 to 1.

**So the given identity cannot be proved.**

We know that the product between the tangent and cotangent functions is 1.

tan x*cot x = 1

We'll raise to square both sides:

(tan x*cot x)^2 = 1

sec x = 1/cos x

(sec x)^2 = 1/(cos x)^2

csc x = 1/ sin x

(csc x)^2 = 1/(sin x)^2

(sec x)^2 + (csc x)^2 = [1/(cos x)^2] + [1/(sin x)^2]

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

(sec x)^2 + (csc x)^2 = 1/(cos x)^2*(sin x)^2

**We notice that the LHS is not equal to RHS, therefore the given expresison does not represent an identity.**