If $x\cos(\theta)-\sin(\theta)=1$ then what is the value of $x^2+(1+x^2)\sin(\theta)=1$ The question given is,

If $x\cos(\theta)-\sin(\theta)=1$ then find the value of $x^2+(1+x^2)\sin(\theta)$.

There are four options given $1$, $-1$, $0$ and $2$. I tried using $\sin^2+\cos^2=1$. I also tried to isolate $x$ and put its value in the second equation but things didn't get simplified.
Please explain how should I solve this question.

The original claim $x^2+(1+x^2)\sin\theta=1$ is false. When $\sin(\theta)=-1$, and $\cos(\theta)=0$ (say at $\theta=-\pi/2$), then for any $x$, we have $x\cos(\theta)-\sin\theta=1$ but
$$
x^2+(1+x^2)\sin\theta=x^2-(1+x^2)=-1\neq 1.
$$

Edit for the updated question: The case $\cos\theta=0$ gives us $\sin(\theta)=-1$ and so
$$
x^2+(1+x^2)\sin\theta=-1.
$$
When $\cos\theta\neq 0$, we have $x\cos\theta-\sin\theta=1\implies x=\frac{1+\sin\theta}{\cos\theta}$ and so
$$
x^2=\frac{1+2\sin\theta+\sin^2\theta}{\cos^2\theta},\quad 1+x^2=\frac{2+2\sin\theta}{\cos^2\theta}
$$
which implies
$$
x^2+(1+x^2)\sin\theta=\frac{1+2\sin\theta+\sin^2\theta+(2\sin\theta+2\sin^2\theta)}{\cos^2\theta}=\frac{2(1+2\sin\theta+\sin^2\theta)+\sin^2\theta-1}{\cos^2\theta}=-1+2x^2.
$$
This means your expression is in general not identically equal to any constant. 
