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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

Consider the following two statements P and Q. P: ^ -1 (cos...

Consider the following two statements PP and Q.Q.
P:cos⁡−1(cos4π3)=4π3P: \cos^{-1} \left(cos \dfrac{4\pi}{3}\right) = \dfrac{4\pi}{3}
Q:sec⁡2(cot⁡−112)+cosec2(tan⁡−113)=15Q: \sec^{2} \left(\cot^{-1} \dfrac{1}{2}\right) + {cosec}^{2}\left(\tan^{-1} \dfrac{1}{3}\right) = 15
Then, which of the following is true?

Options

  1. ABoth PP and QQ are true
  2. BPP is true, but QQ is false
  3. CPP is false, but QQ is true
  4. DBoth PP and QQ are false

Correct answer

C. PP is false, but QQ is true

Step-by-step solution

We have, cos⁡(4π3)=−cosπ3=−12\cos \left(\dfrac{4\pi}{3}\right) = - cos \dfrac{\pi}{3} = - \dfrac{1}{2}

Hence, cos⁡−1(−12)=2π3\cos^{- 1} \left(- \dfrac{1}{2}\right) = \dfrac{2\pi}{3}
Therefore, PP is false.

Now, sec⁡2(cot⁡−112)+cosec2(tan⁡−113)\sec^{2} \left(\cot^{- 1} \dfrac{1}{2}\right) + cose{c}^{2}\left(\tan^{- 1} \dfrac{1}{3}\right)

=[1+tan⁡2(tan⁡−12)]+[1+cot⁡2(cot⁡−13)]= \left[1 + \tan^{2} \left(\tan^{- 1} 2\right)\right] + \left[1 + \cot^{2} \left(\cot^{- 1} 3\right)\right]
=1+4+1+9=15= 1 + 4 + 1 + 9 = 15
Therefore, QQ is true.

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