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

NDA 2024 · Mathematics · Inverse Trigonometric Functions

If 4 ^ -1 x+ ^ -1 x=π , then what is ^ -1 x+4 ^ -1 x equal to?

If 4sin⁡−1x+cos⁡−1x=π4 \sin ^{-1} x+\cos ^{-1} x=\pi, then what is sin⁡−1x+4cos⁡−1x\sin ^{-1} x+4 \cos ^{-1} x equal to?

Options

  1. Aπ2\frac{\pi}{2}
  2. Bπ\pi
  3. C3π2\frac{3 \pi}{2}
  4. D2π2 \pi

Correct answer

C. 3π2\frac{3 \pi}{2}

Step-by-step solution

(c) We have given, 4sin⁡−1x+cos⁡−1x=π4 \sin ^{-1} x+\cos ^{-1} x=\pi. ⇒4(π2−cos⁡−1x)+(π2−sin⁡−1x)=π{sin⁡−1x+cos⁡−1x=π2}\begin{aligned} & \Rightarrow 4\left(\frac{\pi}{2}-\cos ^{-1} x\right)+\left(\frac{\pi}{2}-\sin ^{-1} x\right)=\pi \\ & \left\{\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right\} \end{aligned}⇒2π−4cos⁡−1x+π2−sin⁡−1x=π⇒4cos⁡−1x+sin⁡−1x=3π2\begin{aligned} & \Rightarrow \quad 2 \pi-4 \cos ^{-1} x+\frac{\pi}{2}-\sin ^{-1} x=\pi \\ & \Rightarrow \quad 4 \cos ^{-1} x+\sin ^{-1} x=\frac{3 \pi}{2}\end{aligned}$
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