Question

If $xcos\theta -ysin\theta =a$ and $xsin\theta +ycos\theta =b,$ then $\frac{x^2+y^2}{a^2+b^2}$ = ___.

• A. 2
• B. 1
• C. 0
• D. -1

Answer 1

The correct option is B

1

$xcos\theta -ysin\theta =a$ squaring both sides we get,

$\therefore (xcos\theta -ysin\theta {)}^2=a^2$

$xsin\theta +ycos\theta =b$

$(xsin\theta +ycos\theta {)}^2=b^2$

$\therefore x^{2}co{s}^2\theta -2\times ycos\theta sin\theta +y^{2}si{n}^2\theta =a^2$

Adding, $\frac{x^{2}si{n}^2\theta +2xysin\theta cos\theta +y^{2}co{s}^2\theta =b^2}{x^2(co{s}^2\theta +si{n}^2\theta )+y^2(si{n}^2\theta +co{s}^2\theta )=a^2+b^2}$

$\therefore x^2+y^2=a^2+b^2$ $....(si{n}^2\theta +co{s}^2\theta =1)$

$\therefore \frac{x^2+y^2}{a^2+b^2}=1$