Question

If ${\left({\sin }^{-1}a\right)}^2+{\left({\cos }^{-1}b\right)}^2+{\left({\sec }^{-1}c\right)}^2+{\left({\mathrm{cosec}}^{-1}d\right)}^2=\frac{5{\pi }^2}{2}$ then the value of ${\left({\sin }^{-1}a\right)}^2-{\left({\cos }^{-1}b\right)}^2+{\left({\sec }^{-1}c\right)}^2-{\left({\mathrm{cosec}}^{-1}d\right)}^2$ is equal to

• A. $-{\pi }^2$
• B. $-\frac{{\pi }^2}{2}$
• C. 0
• D. $\frac{{\pi }^2}{2}$

Answer 1

Answer: (C)

As $0\le {\left({\sin }^{-1}a\right)}^2\le \frac{{\pi }^2}{4},0\le {\left({\cos }^{-1}b\right)}^2\le {\pi }^2,0\le {\left({\sec }^{-1}c\right)}^2\le {\pi }^2($ except $\frac{{\pi }^2}{4})$ and $0<{\left({\mathrm{cosec}}^{-1}d\right)}^2\le \frac{{\pi }^2}{4}$
So $0<{\left({\sin }^{-1}a\right)}^2+{\left({\cos }^{-1}b\right)}^2+{\left({\sec }^{-1}c\right)}^2+{\left({\mathrm{cosec}}^{-1}d\right)}^2\le \frac{5{\pi }^2}{2}$
$\therefore {\left({\sin }^{-1}a\right)}^2+{\left({\cos }^{-1}b\right)}^2+{\left({\sec }^{-1}c\right)}^2+{\left({\mathrm{cosec}}^{-1}d\right)}^2=\frac{5{\pi }^2}{2}($ Given $)$
$\Rightarrow {\left({\sin }^{-1}a\right)}^2=\frac{{\pi }^2}{4},{\left({\cos }^{-1}b\right)}^2={\pi }^2,{\left({\sec }^{-1}c\right)}^2={\pi }^2$ and ${\left({\mathrm{cosec}}^{-1}d\right)}^2=\frac{{\pi }^2}{4}$
Hence ${\left({\sin }^{-1}a\right)}^2-{\left({\cos }^{-1}b\right)}^2+{\left({\sec }^{-1}b\right)}^2-{\left({\mathrm{cosec}}^{-1}d\right)}^2=0$