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.
 A: 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. 
A: Let $x=r\sin y,1=r\cos y$ where $r>0$
$\implies r=\sqrt{1+x^2}\implies\cos y=\dfrac1r=\cdots$
$\implies r(\sin y\cos\theta-\cos y\sin\theta)=r\cos y$
$\implies\sin(y-\theta)=\cos y=\sin\left(\dfrac\pi2-y\right)$
$\implies y-\theta=n\pi+(-1)^n\left(\dfrac\pi2-y\right)$ where $n$ is any integer 
If $n$ is even $=-2m,$
$\implies y-\theta=2m\pi+\left(\dfrac\pi2-y\right)\implies\theta=2m\pi+2y-\dfrac\pi2$ 
$\sin\theta=\cdots=-\cos2y=?$
If $n$ is odd $=2m+1$
and so on
A: WLOG let $x=\cot y$
$$\implies\cos(y+\theta)=\sin y=\cos\left(\dfrac\pi2-y\right)$$
$$\implies y+\theta=2m\pi\pm\left(\dfrac\pi2-y\right)$$ where $m$ is any integer
$+\implies\theta=2m\pi+\dfrac\pi2-2y\implies\sin\theta=\cdots=\cos2y=\dfrac{\cot^2y-1}{\cot^2y+1}=?$
Check for the '-' sign
A: From the first equation we obtain $$x=\frac{1+\sin \theta}{\cos \theta}$$
Then $$x^2=\frac {(1+\sin \theta)^2}{\cos^2\theta}=\frac {(1+\sin \theta)^2}{1-\sin^2\theta}=\frac {1+\sin \theta}{1-\sin\theta}$$
That is provided we can cancel $1+\sin \theta$ i.e. $\sin \theta \neq -1$
Clearing fractions this gives $$x^2(1-\sin\theta)=1+\sin \theta$$ or $$1=x^2-(1+x^2)\sin \theta$$
Which wasn't what you wanted to prove, but would seem to be the correct formulation.
