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JEE Main 202225 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2022 · 25 June · Evening Shift · Mathematics · Inverse Trigonometric Functions

The value of ^ - 1 [(cos ((15π)/(4)) - 1)/( ((π)/(4)))] is equal to

The value of tan⁡−1\tan^{- 1}\left[\dfrac{cos\left(\dfrac{15\pi}{4}\right) - 1}{\sin \left(\dfrac{\pi}{4}\right)}\right] is equal to

Options

  1. A−π4- \dfrac{\pi}{4}
  2. B−π8- \dfrac{\pi}{8}
  3. C−5π12- \dfrac{5\pi}{12}
  4. D−4π9- \dfrac{4\pi}{9}

Correct answer

B. −π8- \dfrac{\pi}{8}

Step-by-step solution

Given,

tan⁡−1(cos⁡(15π4)−1sinπ4)\tan^{- 1}\left(\dfrac{\cos \left(\dfrac{15\pi}{4}\right) - 1}{sin\dfrac{\pi}{4}}\right)

=tan⁡−1(cos⁡(4π−π4)−1sinπ4)= \tan^{- 1}\left(\dfrac{\cos \left(4\pi - \dfrac{\pi}{4}\right) - 1}{sin\dfrac{\pi}{4}}\right)

=tan⁡−1(12−112)= \tan^{- 1}\left(\dfrac{\dfrac{1}{\sqrt{2}} - 1}{\dfrac{1}{\sqrt{2}}}\right)

=tan⁡−1(1−2)=−tan⁡−1(2−1)= \tan^{- 1}\left(1 - \sqrt{2}\right) = - \tan^{- 1}\left(\sqrt{2} - 1\right)

=−π8= - \dfrac{\pi}{8}

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