tan x= 5/12

we need to find cos x using the right angled triangle.

We know that in the right angled triangle we can obtain measurements through trigonometric equation:

For the angle x:

tanx= sinx/cosx = opposite side/adjacent side= 5/12

that means that 5 and 12 are the opposite and adjacent sides in a right angle triangle.

Then the Hypotenuse (h) = sqrt[(5)^2 +(12)^2]= 13

Now sinx = opposite side/ hypotenuse = 5/13

cos x= adjacent/hypotenuse = 12/13

tanx = 5/12.To find cosx'

Solution:

We now that cosx = tanx/ sqrt(1+tan^2x) = (5/12)/sqrt[1+(5/12)^2}] = (5/12)/sqrt{1+25/144) = (5*12/12)/sqrt(144+25) = 5/13.

The fundamental formula in trigonometry says that:

sin^2 x+cos^2 x=1

If you divide the above formula with cos^2 x

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

But you know the fact that the tangent function is the ratio between sin x/cos x, so (sin^2 x)/(cos^2 x) = tg^2 x

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

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

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

cos x = [1/(tg^2 x + 1)]^1/2

cos x = {[1/[(5/12)^2 + 1])}^1/2

cos x = {[1/[(25/144) + 1])}^1/2

cos x = [1/(169/144)]^1/2

cos x = (144/169)^1/2

**cos x = 12/13**