We have to verify that (1+ (tan x)^2) *(cos x)^2 = 1

Starting with the left hand side

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

tan x = sin x / cos x

=> (1 + (sin 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

=> 1

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

R:H:S = (1+tan²x).cos²x

= cos²x + tan²x.cos²x

= cos²x + (sin²x/cos²x)cos²x

= cos²x + sin²x

= 1

= L:H:S

We notice that inside brackets we have a consequence of Pythagorean identity:

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

We'll show how it works:

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/(cos 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**