Question

If $xsi{n}^3\theta +yco{s}^3\theta =sin\theta cos\theta$ and $xsin\theta =ycos\theta ,$ then $x^2+y^2=$ ___.

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

The correct option is C

1

Given: $xsi{n}^3\theta +yco{s}^3\theta =sin\theta cos\theta$
$\Rightarrow xsin\theta \left(si{n}^2\theta \right)+yco{s}^3\theta =sin\theta cos\theta .....\left(i\right)$

We know, $xsin\theta =ycos\theta$... (given)
Substitute in (i), we get,
$\therefore ycos\theta si{n}^2\theta +yco{s}^3\theta =sin\theta cos\theta$
$\Rightarrow ycos\theta (si{n}^2\theta +co{s}^2\theta )=sin\theta cos\theta$
$\therefore ycos\theta =sin\theta cos\theta \Rightarrow y=sin\theta ....\left(ii\right)$

Again, from (i),
$xsi{n}^3\theta +yco{s}^3\theta =sin\theta cos\theta$
$xsi{n}^3\theta +ycos\theta \left(co{s}^2\theta \right)=sin\theta cos\theta$

We know, $ycos\theta =xsin\theta$ ....(given)
On substituing, we get,
$\therefore xsi{n}^3\theta +xsin\theta co{s}^2\theta =sin\theta cos\theta$
$xsin\theta (si{n}^2\theta +co{s}^2\theta )=sin\theta cos\theta$
$xsin\theta =sin\theta cos\theta$
$\therefore x=cos\theta ....\left(iii\right)$

Squaring and adding eqn (ii) and (iii) we get,

$x^2+y^2=co{s}^2\theta +si{n}^2\theta$

$\Rightarrow x^2+y^2=1$