# Pythagorean expressionuse Pythagorean expression (cos x)^2*[1 + (tan x)^2] = 1

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### 2 Answers

We have to prove (cos x)^2*[1 + (tan x)^2] = 1

Let's start from the left hand side

(cos x)^2*[1 + (tan x)^2]

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

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

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

cancel (cos x)^2

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

=> 1

**Therefore we prove that (cos x)^2*[1 + (tan x)^2] = 1**

We know, as a consequence of Pythagorean identity, that:

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

Let's see how:

Pythagorean identity states that:

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

We'll divide by (cos x)^2:

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

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

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

Now, we'll substitute what's inside brackets by the equivalent above:** **

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