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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2024 · Mathematics · Inverse Trigonometric Functions

What is ^2 ( ^ -1 2)+ ^2 (cosec^ -1 3) equal to?

What is cot⁡2(sec⁡−12)+tan⁡2(cosec⁡−13)equalto?\cot ^2 \left(\sec ^{-1}2\right)+\tan ^2 \left(\operatorname{cosec}^{-1}3\right) equal to?

Options

  1. A1112\frac{11}{12}
  2. B1124\frac{11}{24}
  3. C724\frac{7}{24}
  4. D124\frac{1}{24}

Correct answer

B. 1124\frac{11}{24}

Step-by-step solution

 Suppose I=cot⁡2(sec⁡−12)+tan⁡2(cosec⁡−13)⇒I=cot⁡2(60∘)+tan⁡2+(cot⁡−19−1){∴cosec⁡−1t=cot⁡−1t2−1}⇒I=13+tan⁡2(tan⁡−118)=13+18=1124\begin{aligned} & \text{ Suppose } \mathrm{I}=\cot ^2\left(\sec ^{-1} 2\right)+\tan ^2\left(\operatorname{cosec}^{-1} 3\right) \\ & \Rightarrow \mathrm{I}=\cot ^2\left(60^{{}^{\circ}}\right)+\tan ^2+\left(\cot ^{-1} \sqrt{9-1}\right) \\ & \left\{\therefore \operatorname{cosec}^{-1} \mathrm{t}=\cot ^{-1} \sqrt{\mathrm{t}^2-1}\right\} \\ & \Rightarrow \mathrm{I}=\frac{1}{3}+\tan ^2\left(\tan ^{-1} \frac{1}{\sqrt{8}}\right)=\frac{1}{3}+\frac{1}{8}=\frac{11}{24}\end{aligned}
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