Question

Let $A(1,2),B(\text{cosec}\alpha ,-2)$ and $C(2,\sec \beta )$ are $3$ points such that $(OA{)}^2=OB\cdot OC$,($O$ is the origin) then the value of $2{\sin }^2\alpha -{\tan }^2\beta$ is

• A. $0$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is B $2$

Given : $(OA{)}^2=OB\cdot OC\Rightarrow 5=\sqrt{{\text{cosec}}^2\alpha +4}\cdot \sqrt{4+{\sec }^2\beta }$
$\Rightarrow ({\text{cosec}}^2\alpha +4)({\sec }^2\beta +4)=25$
only possible condition is : ${\text{cosec}}^2\alpha +4={\sec }^2\beta +4=5$
$\Rightarrow {\text{cosec}}^2\alpha ={\sec }^2\beta =1$
${\sin }^2\alpha ={\cos }^2\beta =1$
$\therefore 2{\sin }^2\alpha -{\tan }^2\beta =2$