Question

If a cosƟ + b sinƟ = and a sinƟ – b cosƟ =y, Prove that a²+b² = ²+² evantell me how to mark this anwr as brainiest i am not able to et the option for it whoever will answer be marksd as braienst

Answer 1

Given that:

a cosƟ + b sinƟ = x

a sinƟ - b cosƟ = y

Squaring on both sides:

a^2 cos^2 Ɵ + b^2 sin^2 Ɵ + 2ab sinƟ cosƟ = x^2

a^2 sin^2 Ɵ + b^2 cos^2 Ɵ - 2ab sinƟ cosƟ = y^2

Find the value of 2ab sinƟ cosƟ from both equations:

2ab sinƟ cosƟ = x^2 - a^2 cos^2 Ɵ - b^2 sin^2 Ɵ

2ab sinƟ cosƟ = a^2 sin^2 Ɵ + b^2 cos^2 Ɵ - y^2

Equate the equations:

x^2 - a^2 cos^2 Ɵ - b^2 sin^2 Ɵ = a^2 sin^2 Ɵ + b^2 cos^2 Ɵ - y^2

Solve it further:

x^2 + y^2 = a^2 sin^2 Ɵ + b^2 cos^2 Ɵ + a^2 cos^2 Ɵ + b^2 sin^2 Ɵ

x^2 + y^2 = a^2(sin^2 Ɵ + cos^2 Ɵ) + b^2 (cos^2 Ɵ + sin^2 Ɵ)

x^2 + y^2 = a^2 + b^2

hence proved.

Identities used:

• (a+b)^2 = a^2 + b^2 + 2ab
• (a-b)^2 = a^2 + b^2 - 2ab
• sin^2 Ɵ + cos^2 Ɵ = 1

Answer 2

Correct Question:

If x = a cos Θ + b sin Θ and y = a sin Θ - b cos Θ , prove that x² +y² = a²+ b².

Your Answer:

Given:-

• x = a cos Θ + b sin Θ
• y = a sin Θ - b cos Θ

To Prove:-

• x² +y² = a²+ b²

Solution:-

Solving LHS

$\tt x^{2}+y^{2}\\\\ \tt = (a \cos \theta + b \sin \theta)^{2}+ (a \sin \theta - b \cos \theta)^{2}\\\\ \tt =(a \cos\theta)^{2}+(b\sin\theta)^{2}+2ab\sin\theta\cos\theta+(a \sin\theta)^{2}+(b\cos\theta)^{2}-2ab\sin\theta\cos\theta\\\\ \tt =a^{2}{\cos}^{2}\theta+b^{2}{\sin}^{2}\theta+a^{2}{\sin}^{2}\theta+b^{2}{\cos}^{2}\theta\\\\ \tt = a^{2}({\cos}^{2}\theta+{\sin}^{2}\theta)+b^{2}({\cos}^{2}\theta+{\sin}^{2}\theta)\\\\ = \tt a^{2}+b^{2}\:\:\:\:\:\:\:\:\:\:(\because {\cos}^{2}\theta+{\sin}^{2}\theta=1)$

LHS= RHS

proved