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. $x^2-2x=sin2\theta$
• B. $y^2+2y=sin2\theta$
• C. $xy=sin2\theta$
• D. $x-y=2$

Answer 1

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

The correct options are
A $y^2+2y=sin2\theta$
B $x^2-2x=sin2\theta$
C $x-y=2$
D $xy=sin2\theta$

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

$\Rightarrow sin\theta +cos\theta =x-1$ ...(1)

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

$\Rightarrow sin\theta -{sin}^2\theta +cos\theta -{cos}^2\theta =y$

$\Rightarrow sin\theta +cos\theta =y+1$ ...(2)

Equating (1) and (2)

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

Now as $sin2\theta ={(sin\theta +cos\theta )}^2-1$ ...(3)

Substituting (1) in (3), we get

$sin2\theta ={(x-1)}^2-1=x^2+1-2x-1=x^2-2x$

Substituting (2) in (3), we get

$sin2\theta ={(y+1)}^2-1=y^2+1+2y-1=y^2+2y$

And From $\left(1\right),\left(2\right)$

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

$xy={(sin\theta +cos\theta )}^2-1\Rightarrow xy=sin2\theta$

Hence, all options are correct.