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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2025 · Mathematics · Inverse Trigonometric Functions

What is ^ -1 9 + cosec^ -1 ( 41 4 ) equal to?

What is cot⁡−19+cosec−1(414)equalto?\cot^{-1} 9 + \text{cosec}^{-1} \left(\dfrac{\sqrt{41}}{4}\right) equal to?

Options

  1. Aπ4\dfrac{\pi}{4}
  2. Bπ3\dfrac{\pi}{3}
  3. Cπ2\dfrac{\pi}{2}
  4. Dπ\pi

Correct answer

A. π4\dfrac{\pi}{4}

Step-by-step solution

Let A=cot⁡−19⇒cot⁡A=9⇒tan⁡A=19A = \cot^{-1} 9 \Rightarrow \cot A = 9 \Rightarrow \tan A = \dfrac{1}{9}

Let B=cosec−1(414)⇒cosecB=414⇒sin⁡B=441B = \text{cosec}^{-1}\left(\dfrac{\sqrt{41}}{4}\right) \Rightarrow \text{cosec} B = \dfrac{\sqrt{41}}{4} \Rightarrow \sin B = \dfrac{4}{\sqrt{41}}

From this, tan⁡B=4(41)2−42=441−16=45\tan B = \dfrac{4}{\sqrt{(\sqrt{41})^2 - 4^2}} = \dfrac{4}{\sqrt{41 - 16}} = \dfrac{4}{5}

Using the formula tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}:

tan⁡(A+B)=19+451−(19)(45)\tan(A + B) = \dfrac{\dfrac{1}{9} + \dfrac{4}{5}}{1 - \left(\dfrac{1}{9}\right)\left(\dfrac{4}{5}\right)}

tan⁡(A+B)=5+364545−445\tan(A + B) = \dfrac{\dfrac{5 + 36}{45}}{\dfrac{45 - 4}{45}}

tan⁡(A+B)=4141=1\tan(A + B) = \dfrac{41}{41} = 1

Since A,B∈(0,π2),wehaveA, B \in \left(0, \dfrac{\pi}{2}\right), we haveA + B = π4\dfrac{\pi}{4}<br><br>Answer:π4<br><br>Answer: \dfrac{\pi}{4}
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