# Given cot x=1/5, what is the value cos^2 x?

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### 3 Answers

We have cot x= 1/5. We need the value of (cos x)^2

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

divide all terms by (sin x)^2

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

substitute cot x = 1/5

=> 1 + (1/5)^2 = 1/(sin x)^2

=> 1/(sin x)^2 = 1 + 1/25

=> 1/(sin x)^2 = 26/25

=> (sin x)^2 = 25/26

=> 1 - (cos x)^2 = (25/26)

=> (cos x)^2 = (1 - 25/26)

=> (cos x)^2 = (1/26)

**The value of (cos x)^2 = 1/26**

cot x= 1/tan x

therefore,

since 1/5= 1/tan x, tan x =5

In a right angled triangle, tan x= opposite side/ adjacent side = 5/1=5

so, hypo. side= square root 26

cos x = adjacent side / hypo. side= 1/ square root 26

therefore since cos 2x= 1- 2cos^2 x= 1- 2(1 / square root 26) ^2

If cot x = 1/5, then tan x = 1/cot x = 5.

But tan x = sin x/cos x => sin x/cos x = 5

sin x = 5*cos x

We'll raise to square both sides:

(sin x)^2 = 25*(cos x)^2

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

(sin x)^2 + (cos x)^2 = 25*(cos x)^2 + (cos x)^2

But, from Pythagorean identity, we'll have:

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

1 = 26*(cos x)^2

We'll divide by 26 and we'll use symmetric property:

(cos x)^2 = 1/26

**The value of (cos x)^2 is (cos x)^2 = 1/26.**