Question

The number of solutions $(x,y)$ of the equation $1+x^2+2x\sin \left({\cos }^{-1}y\right)=0,$ are

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (D)

$1$

For given equation to hold true when $x=1,y=0$

So, Putting these values in the Equation :- $1+x^2+2x\sin \left({\cos }^{-1}y\right)=0$

$\Rightarrow 1+1^2-2\ast 1\ast \sin \left({\cos }^{-1}\right(0\left)\right)=1+1^2-2\ast 1\ast \sin \left(\frac{3\pi }{2}\right)=1+1-2=0$

Hence, only $1$ solution is possible