` tan x = -4/3 , x ` द्वितीय चतुर्थाश में है ।

Explanation

` :' tan x = -4/3 `
` :. (2 tan .x/2 )/( 1 - tan ^(2).x/2) = -4/3`
या ` - 4 + 4 tan^(2) . x/2 = 6 tan. x/2 `
या ` 4 tan^(2) .x/6 - 6 tan.x/2 - 4 = 0 ,`
या ` 4 tan^(2) .x/6 - 6 tan. x/2 - 4 = 0 ,`
` 2 tan^(2) .x/2 - 3 tan . x/2 - 2 = 0 ,`
` 2 tan^(2) . x/2 - 4 tan. x/2 + tan.x/2 -2 = 0,`
` 2 tan^(2).x/2 (tan.x/2 - 2) + 1 ( tan. x/2 - 2) = 0 `
` (2 tan .x/2 +1)(tan .x/2 - 2)=0`
`( 2 tan . x/2 +1) = 0 " " (tan .x/2 - 2)=0`
` 2 tan. x/2 + 1 = 0 " " tan. x/2 = -1/2 `
` tan. x/2 - 2 = 0 " " tan. x/2 = 2 `
परन्तु x द्वितीय चतुर्थाश में है जिसमे sin x धनात्मक होगा । अतः `tan.x/2` का मान`-1/2` अमान्य है ।
तब ` tan. x/2 = 2`
तो ` cosx = (1- tan^(2) .x/2 )/(1+tan^(2).x/2)=(1-(2)^(3))/(1+(2)^(2))= -3/5`
` : ' cos x = -3/5`
` rArr 1 - 2 sin^(2) .x/2 = -3/5 "अथवा " 2 cos^(2) . x/2 -1 =- 3/5`
यदि ` 1- 2 sin^(2). x/2 = -3/5`
तो ` 2 sin^(2) . x/2 = 1 + 3/5 = 8/5 "या " sin^(2) .x/2 = 4/5 `
या ` sin .x/2 = 2/(sqrt(2))`
यदि ` 2 cos^(2) . x/2 -1 = -3/5`
तो ` 2 cos^(2) . x/2 = 1 - 3/5 = 2/5 " या " cos^(2) . x/2 = 1/5`
या` cos . x/2 = 1/(sqrt(5))`
अतः ` sin . x/2 = 2 / (sqrt(5)), cos.x/2 = 1/ (sqrt(5)) " तथा " tan .x/2 = 2 `