Question

if xcos theta-ysin theta =a, xsin theta+ycos theta=b, prove that x²+y²=a²+b²

Answer 1

Given-
Xcosø-Ysinø= a
Xsinø+Ycosø= b
To prove-
X²+Y²=a²+b²

Proof-
a²+ b²
= (Xcosø-Ysinø)² +(Xsinø+Ycosø)²
=X²cos²ø-2XcosøYsinø+Y²sin²ø+X²sin²ø+2XsinøYcosø+Y²cos²ø
=X²(cos²ø+sin²ø) +Y²(sin²ø+cos²ø)+0
=X²+Y²
Hence proved,
a²+b²=X²+Y²

Answer 2

$x^{2}+y^{2} =a^{2} +b^{2}$, proved.

Step-by-step explanation:

We have,

$x\cos \theta -y\sin \theta=a$     ....(1)

and ,

$x\sin \theta +y\cos \theta=b$    ... (2)

Squaring and adding (1) and (2), we get

$(x\cos \theta -y\sin \theta)^{2} + (x\sin \theta +y\cos \theta)^{2} =a^{2} +b^{2}$

⇒$(x^{2} \cos^{2} \theta+y^{2} \sin^{2} \theta-2xy\sin \theta\cos \theta )+(x^{2} \cos^{2} \theta+y^{2} \sin^{2} \theta+2xy\sin \theta\cos \theta )=a^{2} +b^{2}$

⇒ $x^{2} \cos^{2} \theta+y^{2} \sin^{2} \theta+x^{2} \cos^{2} \theta+y^{2} \sin^{2} \theta=a^{2} +b^{2}$

⇒ $x^{2}( \cos^{2} \theta+ \sin^{2} \theta)+y^{2} (\cos^{2} \theta+\sin^{2} \theta)=a^{2} +b^{2}$

⇒$x^{2}(1)+y^{2} (1)=a^{2} +b^{2}$

⇒ $x^{2}+y^{2} =a^{2} +b^{2}$, proved.