# What is cos^2x if tanx=4?

*print*Print*list*Cite

### 2 Answers

We have tan x = 4.

Now we have to find (cos x)^2.

tan x = sin x / cos x

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

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

=> (sin x)^2 = 1 - (cos x)^2

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

=> (tan x)^2 = [1 - (cos x)^2] / (cos x)^2 = 4^2 = 16

=> [1 - (cos x)^2] = 16*(cos x)^2

=> 1 = 17*(cos x)^2

=> (cos x)^2 = 1/17.

**Therefore (cos x)^2 = 1/17.**

We'll start from the fundamental formula of trigonometry:

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

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

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

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

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

From enunciation, we know that tan x = 4.

We'll square raise both sides:

(tan x)^2 = 16

16 + 1 = 1/(cos x)^2

1/(cos x)^2 = 17

**(cos x)^2 = 1/17**