Question

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

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

Answer 1

Answer: (C)

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