# How to prove the identity sin^4x - cos^4x / sin^3x - cos^3x = Sinx + cosx / 1 + sinxcosx

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to prove that: [(sin x)^4 - (cos x)^4]/[(sin x)^3 - (cos x)^3 = (sin x + cos x) / (1 + sin x* cos x)

[(sin x)^4 - (cos x)^4]/[(sin x)^3 - (cos x)^3

use a^2 - b^2 = (a - b)(a + b) and a^3 - b^3 = (a - b)*(b^2  + a*b + a^2)

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

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

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

=> (sin x + cos x)*1/(sin x - cos x)(1 + sin x*cos x)

This proves: [(sin x)^4 - (cos x)^4]/[(sin x)^3 - (cos x)^3 = (sin x + cos x) / (1 + sin x* cos x)

lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

L:H:S = (sin⁴x - cos⁴x)/(sin³x - cos³x)

⇒ x² - y² = (x + y)(x - y)

x³ - y³ = (x - y)(x² + y² + xy)

= (sin²x-cos²x)(sin²x+cos²x)/(sinx-cosx)(sin²x+cos²x+sinx.cosx)

we know that sin²x+cos²x=1

= (sin²x-cos²x)/(sinx-cosx)(1+sinx.cosx)

= (sinx-cosx)(sinx+cosx)/(sinx-cosx)(1+sinx.cosx)

= (sinx+cosx)/(1+sinx.cosx)

= R:H:S