JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of \(16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right)\) is :
- A \(24 \pi^2\)
- B \(22 \pi^2\)
- C \(31 \pi^2\)
- D \(18 \pi^2\)
Answer & Solution
Correct Answer
(B) \(22 \pi^2\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & 16\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2 \\ & \operatorname{Sec}^{-1} x=a \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \\ & \operatorname{cosec}^{-1} x=\frac{\pi}{2}-a \\ & =16\left[a^2+\left(\frac{\pi}{2}-a\right)^2\right]=16\left[2…
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