Question 125

If $$x=a\ \cos\theta+b\ \sin\theta$$ and $$y=a\ \sin\theta-b\ \cos\theta$$, the value of $$x^2+y^2$$ is:

Solution

$$x=aCos\theta + bSin\theta $$

$$x^2=(aCos\theta +bSin\theta)^2$$

$$=a^2Cos^2\theta+b^Sin^2\theta+2ab$$

$$Cos\theta Sin\theta $$.

Similarly,

$$y^2=b^2Cos^2\theta +a^2Sin^2\theta$$

$$-2abCos\theta Sin\theta$$.

So,$$x^2+y^2$$

$$=(a^2+b^2)Cos^2\theta + (a^2+b^2)$$

$$Sin^2\theta$$.

$$=(a^2+b^2)(Cos^2\theta+Sin^2\theta).$$

we know,$$Cos^2\theta +Sin^2\theta=1$$.

So,$$x^2+y^2=a^2+b^2.$$

Option C is correct.

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