We have to find cos x, if x is in interval (pi/2 ,pi) and tan x=-12/5

tan x = -12/5

=> sin x/cos x = -12/5

let y = cos x

=> sqrt(1 - y^2)/y = -12/5

=> 5*sqrt (1 - y^2) = -12*y

square both the sides

=> 25(1 - y^2) = 144y^2

=> 25 - 15y^2 = 144y^2

=> 25 = 169y^2

=> y^2 = 25/169

y = 5/13 and y = -5/13

cos x = 5/13 and cos x = -5/13

But as x lies in (pi/2, pi) we eliminate cos x = 5/13

**The value of cos x = -5/13**

We'll apply Pythagorean identity:

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

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

We'll plug in the given value of tan x:

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

(cos x)^2 = 1/(144/25 + 1)

(cos x)^2 = 25/(144+25)

(cos x)^2 = 25/169

cos x = +sqrt (25/169) or cos x = -sqrt (25/169)

Since the values of cosine function are negative in the second quadrant (`pi` /2 ; `pi` ), we'll keep only the negative value for cos x.

cos x = - 5/13

**The requested value for cos x is: cos x = - 5/13.**