Question

If $\sin \theta (1+\sin \theta )+\cos \theta (1+\cos \theta )=x$ and $\sin \theta (1-\sin \theta )+\cos \theta (1-\cos \theta )=y$, then

• A. $x^2-2x-\sin 2\theta =0$
• B. $y^2+2y-\sin 2\theta =0$
• C. $xy=\sin 2\theta$
• D. $xy=\sin \theta +\cos \theta$

Answer 1

Answer: (A), (B), (C)

$xy=\sin 2\theta$
$x^2-2x-\sin 2\theta =0$
$y^2+2y-\sin 2\theta =0$

Given equations are
$\sin \theta (1+\sin \theta )+\cos \theta (1+\cos \theta )=x$

$\sin \theta (1-\sin \theta )+\cos \theta (1-\cos \theta )=y$

On solving both equations, we get
$\sin \theta +{\sin }^2\theta +\cos \theta +{\cos }^2\theta =x$
$\sin \theta +\cos \theta +1=x\to \left(1\right)$

Similarly
$\sin \theta (1-\sin \theta )+\cos \theta (1-\cos \theta )=y$
$\sin \theta -{\sin }^2\theta +\cos \theta -{\cos }^2\theta =y$
$\sin \theta +\cos \theta -1=y\to \left(2\right)$

On multiplying equation (1) and (2), we get

$(\sin \theta +\cos \theta +1)(\sin \theta +\cos \theta -1)=xy$

${(\sin \theta +\cos \theta )}^2-1=xy$
${\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta -1=xy$
$1+2\sin \theta \cos \theta -1=xy$
$\sin 2\theta =xy$
Option C is correct
Similarly, we can check other options as

$x^2-2x-\sin 2\theta =0$
$x^2-2x=\sin 2\theta$

Taking LHS , we have

$x^2-2x$
$=(\sin \theta +\cos \theta +1{)}^2-2(\sin \theta +\cos \theta +1)$
$=(\sin \theta +cos\theta +1)(\sin \theta +\cos \theta +1-2)$
$={(\sin \theta +\cos \theta )}^2-1$
$={\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta -1$
$=1+2\sin \theta \cos \theta -1$
$=\sin 2\theta$
$=RHS$

Similarly, we can prove other option
$y^2+2y-\sin 2\theta =0$
$y^2+2y=\sin 2\theta$
Taking LHS
$=y^2+2y$
$=y(y+2)$
$=(\sin \theta +\cos \theta -1)(\sin \theta +\cos \theta -1+2)$
$={(\sin \theta +\cos \theta )}^2-1$
$={\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta -1$
$=1+2\sin \theta \cos \theta -1$
$=\sin 2\theta$
$=RHS$

Hence, Option A , B and C are correct.