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

NDA 2025 · Mathematics · Inverse Trigonometric Functions

If ^ -1 k + ^ -1 (1)/(2) = (π)/(4) , then what is the value of k ?

If tan⁡−1k+tan⁡−112=π4\tan^{-1}k + \tan^{-1}\dfrac{1}{2} = \dfrac{\pi}{4}, then what is the value of kk?

Options

  1. A11
  2. B12\dfrac{1}{2}
  3. C13\dfrac{1}{3}
  4. D14\dfrac{1}{4}

Correct answer

C. 13\dfrac{1}{3}

Step-by-step solution

Given tan⁡−1k+tan⁡−112=π4\tan^{-1}k + \tan^{-1}\dfrac{1}{2} = \dfrac{\pi}{4}

Using the formula tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy),weget:<br><br>tan⁡−1(k+121−k2)=π4<br><br>2k+12−k=tan⁡(π4)<br><br>2k+12−k=1\tan^{-1}x + \tan^{-1}y = \tan^{-1} \left(\dfrac{x+y}{1-xy}\right), we get:<br><br>\tan^{-1}\left(\dfrac{k + \dfrac{1}{2}}{1 - \dfrac{k}{2}}\right) = \dfrac{\pi}{4}<br><br>\dfrac{2k + 1}{2 - k} = \tan \left(\dfrac{\pi}{4}\right)<br><br>\dfrac{2k + 1}{2 - k} =1

2k+1=2−k2k + 1 = 2 - k

3k=13k = 1

k=13k = \dfrac{1}{3}

Answer: 13\dfrac{1}{3}
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