Question

x= a cos + b sin and y = a sin - b cos , prove that x^2+y^2= a^2+b^2​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{x=a\,cos(\theta)+b\,sin(\theta)\,\,\,\,\,and\,\,\,\,\,y=a\,sin(\theta)-b\,cos(\theta)}$

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

Add the above expressions

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

$\tt{\implies\,x^2+y^2=a^2\left\{cos^2(\theta)+sin^2(\theta)\right\}+b^2\left\{sin^2(\theta)+cos^2(\theta)\right\}}$

$\tt{\implies\,x^2+y^2=a^2\left\{1\right\}+b^2\left\{1\right\}}$

$\tt{\implies\,x^2+y^2=a^2+b^2}$

Answer 2

Step-by-step explanation:

I hope this will be helpful to you and rate the answer

if u r satisfied with the answer mark me as brainliest