Question

Cos^3a+sin^3a/cosa+sina+cos3a-sin3a/cosa-sina=2

Answer 1

LHS:-

$=\bold{\frac{(cosA+sinA)(cos^2+sin^2a-sinA.cosA)}{cosA+sinA} +\frac{(cosA-sinA)(sin^2A+cos^2A+sinA.cosA)}{cosA-sinA} }$

$=\bold{cos^2A+sin^2A-sinA.cosA+sin^2A+cos^2A-sinA.cosA}\\=\bold{1+1}\\=\bold{2=RHS}$

Answer 2

Concept:

Trigonometric Identities are equality statements that hold true for all values of the variables in the equation and that use trigonometry functions.

There are numerous distinctive trigonometric identities that relate a triangle's side length and angle. Only the right-angle triangle is consistent with the trigonometric identities.

The six trigonometric ratios serve as the foundation for all trigonometric identities. Sine, cosine, tangent, cosecant, secant, and cotangent are some of their names. The adjacent side, opposite side, and hypotenuse side of the right triangle are used to define each of these trigonometric ratios. The six trigonometric ratios are the source of all fundamental trigonometric identities.

sin x,cos x , tan x , sec x , cot x, cosec x

sin² A +cos²A=1

cosec² A- cot²A=1

sec² A - tan²A=1

Given:

(cos³A+sin³a)/(sin A +cosA) +(cos³A-sin³a)/(sin A -cosA)=2

Find:

Prove that,(cos³A+sin³a)/(sin A +cosA) +(cos³A-sin³a)/(sin A -cosA)=2

Solution:

LHS=

(cos³A+sin³a)/(sin A +cosA) +(cos³A-sin³a)/(sin A -cosA)

We know,a³+b³=(a+b)(a²-ab+b²) and

               a³-b³=(a-b)(a²+ab+b²)

So,

(sin A +cosA)((sin² A -sinA cosA+cos²A) /(sin A +cosA) + (sin A +cosA)(sin² A +sinAcosA+cos²A)/(sin A +cosA)

but,sin² A +cos²A=1

=1-sinAcosA +1+sinAcosA

=2=RHS

Hence, proved

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