Question : If $x=\operatorname{cosec \theta}-\sin\theta$ and $y=\sec\theta-\cos\theta$, then the relation between $x$ and $y$ is:

Option 1: $x^{2}+y^{2}+3=1$

Option 2: $x^{2}y^{2}\left ( x^{2}+y^{2}+3 \right )=1$

Option 3: $x^{2}\left ( x^{2}+y^{2}-5 \right )=1$

Option 4: $y^{2}\left ( x^{2}+y^{2}-5 \right )=1$

Correct Answer: $x^{2}y^{2}\left ( x^{2}+y^{2}+3 \right )=1$

Solution : $x=\operatorname{cosec \theta}-\sin\theta$
⇒ $x=\frac{1}{\sin\theta}-\sin\theta=\frac{1-\sin^2\theta} {\sin\theta}=\frac{\cos^2\theta}{\sin\theta}$
Also, $x^2=(\frac{\cos^2\theta}{\sin\theta})^2$_____ (i)
$y=\sec\theta-\cos\theta$
⇒ $y=\frac{1}{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\frac{\sin^2\theta}{\cos\theta}$
⇒ $y^2=(\frac{\sin^2\theta}{\cos\theta})^2$_____ (ii)
So, $x^2+y^2=(\frac{\cos^2\theta}{\sin\theta})^2+(\frac{\sin^2\theta}{\cos\theta})^2$
⇒ $x^2+y^2=\frac{\cos^4\theta}{\sin^2\theta}+\frac{\sin^4\theta}{\cos^2\theta}$
⇒ $x^2+y^2=\frac{\cos^6\theta+\sin^6\theta}{\sin^2\theta\cos^2\theta}$
Add 3 to both sides, we get,
⇒ $x^2+y^2+3=\frac{\cos^6\theta+\sin^6\theta}{\sin^2\theta\cos^2\theta}+3$
⇒ $x^2+y^2+3=\frac{\cos^6\theta+\sin^6\theta+3\sin^2\theta\cos^2\theta(\sin^2\theta+\cos^2\theta)}{\sin^2\theta\cos^2\theta}$
⇒ $x^2+y^2+3=\frac{(\cos^2\theta+\sin^2\theta)^3}{\sin^2\theta\cos^2\theta}$
⇒ $x^2+y^2+3=\frac{1}{\sin^2\theta\cos^2\theta}$
⇒ $x^2+y^2+3=\frac{\sin^2\theta\cos^2\theta}{\sin^4\theta\cos^4\theta}$
⇒ $x^2+y^2+3=\frac{1}{(\frac{\cos^2\theta}{\sin\theta})^2×(\frac{\sin^2\theta}{\cos\theta})^2}$
From equation (i) and (ii), we get,
$x^2+y^2+3=\frac{1}{x^2y^2}$
$\therefore x^{2}y^{2}\left ( x^{2}+y^{2}+3 \right )=1$
Hence, the correct answer is $x^{2}y^{2}\left ( x^{2}+y^{2}+3 \right )=1$.