# Simplify the fraction [﻿cosx*sin(2x)-2sinx]/sin^2 x*cosx=?

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to simplify: [﻿cos x*sin(2x) - 2 sin x]/(sin x)^2 *cos x

[﻿cos x*sin(2x) - 2 sin x]/(sin x)^2 *cos x

use sin 2x = 2*sin x * cos x

=> [2*(﻿cos x)^2 *sin x - 2 sin x]/(sin x)^2 *cos x

=> [2*(﻿cos x)^2 - 2]/sin x *cos x

=> [2*(﻿cos x)^2/ sin x*cos x] - [2/sin x *cos x]

=> [2*cos x/ sin x] - [4/ 2*sin x*cos x]

=> [2*cos x/ sin x] - [4/ sin 2x]

=> 2*cot x - 4*cosec 2x

The required result is 2*cot x - 4*cosec 2x

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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[﻿cosx*sin(2x)-2sinx]/sin^2 x*cosx=(2sin x*(cos x)^2 - 2sin x)/(sin x)^2*cos x

(2sin x*(cos x)^2 - 2sin x)/(sin x)^2*cos x = 2sin x*[(cos x)^2 - 1]/(sin x)^2*cos x

But, from Pythagorean identity, we'll get: (cos x)^2 - 1 = -(sin x)^2

2sin x*[(cos x)^2 - 1]/(sin x)^2*cos x = 2sin x*[ -(sin x)^2]/(sin x)^2*cos x

We'lls implify and we'll get:

2sin x*[ -(sin x)^2]/(sin x)^2*cos x = -2sin x/cos x

2sin x*[ -(sin x)^2]/(sin x)^2*cos x = -2 tan x

[﻿cosx*sin(2x)-2sinx]/(sin x)^2*cosx = -2 tan x

lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

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(cosx.sin2x-2sinx)/sin²x.cosx

=(2sinx.cos²x-2sinx)/sin²x.cosx

=2sinx(cos²x-1)/sin²x.cosx

= -2(1-cos²x)/sinx.cosx

= -2sin²x/sinx.cosx

= -2sinx/cosx

= -2tanx