Question

General solution of the equation ${\sin }^{2}x+{\text{cosec}}^{2}x-2(\sin x+\text{cosec}x)+2=0$ is

• A. $(4n-1)\frac{\pi }{2},n\in Z$
• B. $(2n-1)\frac{\pi }{2},n\in Z$
• C. $(4n+1)\frac{\pi }{2},n\in Z$
• D. $(2n+1)\frac{\pi }{2},n\in Z$

Answer 1

The correct option is C $(4n+1)\frac{\pi }{2},n\in Z$

${\sin }^{2}x+{\text{cosec}}^{2}x-2(\sin x+\text{cosec}x)+2=0$
$\Rightarrow ({\sin }^{2}x-2\sin x+1)+({\text{cosec}}^{2}x-2\text{cosec}x+1)=0$
$\Rightarrow (\sin x-1{)}^2+(\text{cosec}x-1{)}^2=0$
$\Rightarrow \sin x-1=0$ and $\text{cosec}x-1=0$
$\Rightarrow \sin x=1$ and $\text{cosec}x=1$
$\Rightarrow x=(4n+1)\frac{\pi }{2};\text{where}n\in Z$