Question

Prove: sin^3 theta + cos^3 theta / sin theta + cos theta ) + sin theta.cos theta =1

Answer 1

use the identity (a^3+b^3)= (a+b)(a^2+b^2-ab)

sin^3A + cos^3A÷(sinA +cos ). + sinA.cosA
(sinA+cosA)(sin^2+cos^2 -sinA.cosA)÷ (sinA + cosA)+ sinA.cosA

or( sinA^2+cosA^2 -sinA.cosA) +sinA.cosA

1-sinA.cosA+sinA.cosA
1

Answer 2

Given,

Prove that;

⇒ $\frac{sin^3x+cos^3x}{sinx+cosx} + sinx.cosx = 1$

Solution,

We can solve this mathematical problem by using the following mathematical process.

The method to prove the given trigonometric identity is as follows,

Considering the left side of the equation(L.H.S) to prove further;

      ⇒ $\frac{sin^3+cos^3x}{sinx + cosx}+ sinx. cosx$

By applying the formula;

∴(a³+b³)=(a+b)(a²+b²-ab)

       ⇒ $\frac{(sinx+cosx)(sin^2x+cos^2x- sinx . cosx)}{sinx+cosx} + sinx.cosx$

       ⇒ (sin²x+cos²x-sinx*cosx) + sinx.cosx

By the trigonometric identity;

∴sin²x + cos²x = 1

       ⇒ 1 - sinx. cosx + sinx.cosx

       ⇒ 1

(L.H.S = R.H.S)

Thus, we can conclude that the left side of the equation is equal to the right side of the equation. Hence, proved.