Question

prove that which is mentioned in the question
pls help it is urgent

Answer 1

Step-by-step explanation:

Given :-

[(Cos^3 θ + Sin^3 θ)/(Cos θ+Sin θ)]+

[(Cos^3 θ- Sin^3 θ)/(Cos θ- Sin θ)]

To find:-

Prove that :

[(Cos^3 θ + Sin^3 θ)/(Cos θ+Sin θ)]+

[(Cos^3 θ- Sin^3 θ)/(Cos θ- Sin θ)] = 2

Solution:-

Given that

On taking LHS :

[(Cos^3 θ + Sin^3 θ)/(Cos θ+Sin θ)]+

[(Cos^3 θ- Sin^3 θ)/(Cos θ- Sin θ)] -----(1)

[(Cos^3 θ + Sin^3 θ)/(Cos θ+Sin θ)]

Cos^3 θ+ Sin^3 θ is in the form of a^3+b^3

Where a = Cos θ and b= Sin θ

We know that

a^3+b^3 = (a+b)(a^2-ab+b^2)

Cos^3 θ+ Sin^3 θ

=> (Cos θ+ Sin θ)(Cos^2θ+Sin^2θ-Cos θSinθ)

Now ,

[(Cos^3 θ + Sin^3 θ)/(Cos θ+Sin θ)]

=> (Cosθ+Sinθ)(Cos^2θ+Sin^2θ-CosθSinθ)/

(Cos θ +Sinθ)

=> Cos^2θ+Sin^2θ-CosθSinθ

we know that

Sin^2 A + Cos^2 A = 1

=> 1- Cos θ Sin θ ----(2)

and

[(Cos^3 θ- Sin^3 θ)/(Cos θ- Sin θ)]

Cos^3 θ- Sin^3 θ is in the form of a^3-b^3

Where a = Cos θ and b= Sin θ

We know that

a^3-b^3 = (a-b)(a^2+ab+b^2)

Cos^3 θ- Sin^3 θ

=> (Cos θ-Sin θ)(Cos^2θ+Sin^2θ+Cos θSinθ)

Now ,

[(Cos^3 θ - Sin^3 θ)/(Cos θ-Sin θ)]

=> (Cosθ-Sinθ)(Cos^2θ+Sin^2θ+CosθSinθ)/

(Cos θ -Sinθ)

=> Cos^2θ+Sin^2θ CosθSinθ

we know that

Sin^2 A + Cos^2 A = 1

=> 1+Cos θ Sin θ ----(3)

From (2) and (3)

(1) can be written as

=>[(Cos^3 θ + Sin^3 θ)/(Cos θ+Sin θ)]+

[(Cos^3 θ- Sin^3 θ)/(Cos θ- Sin θ)]

=> 1+Cos θ Sin θ+ 1-Cos θ Sin θ

=> (1+1)+(Cos θ Sin θ-Cos θ Sin θ)

=> 2+0

=> 2

=> RHS

LHS = RHS

Hence, Proved.

Used formulae:-

• a^3-b^3 = (a-b)(a^2+ab+b^2)

• a^3+b^3 = (a+b)(a^2-ab+b^2)

• Sin^2 A + Cos^2 A = 1