Q199 / 269 · NDA 2025 (Phase 1) · Mathematics
If $\tan^{-1}k + \tan^{-1}\dfrac{1}{2} = \dfrac{\pi}{4}$, then what is the value of $k$?
- A$1$
- B$\dfrac{1}{2}$
- Correct answerC$\dfrac{1}{3}$
- D$\dfrac{1}{4}$
Solution
Given $\tan^{-1}k + \tan^{-1}\dfrac{1}{2} = \dfrac{\pi}{4}$
Using the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$, we get:
$\tan^{-1}\left(\dfrac{k + \dfrac{1}{2}}{1 - \dfrac{k}{2}}\right) = \dfrac{\pi}{4}$
$\dfrac{2k + 1}{2 - k} = \tan\left(\dfrac{\pi}{4}\right)$
$\dfrac{2k + 1}{2 - k} = 1$
$2k + 1 = 2 - k$
$3k = 1$
$k = \dfrac{1}{3}$
Answer: $\dfrac{1}{3}$
Using the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$, we get:
$\tan^{-1}\left(\dfrac{k + \dfrac{1}{2}}{1 - \dfrac{k}{2}}\right) = \dfrac{\pi}{4}$
$\dfrac{2k + 1}{2 - k} = \tan\left(\dfrac{\pi}{4}\right)$
$\dfrac{2k + 1}{2 - k} = 1$
$2k + 1 = 2 - k$
$3k = 1$
$k = \dfrac{1}{3}$
Answer: $\dfrac{1}{3}$
Hence, the correct answer is option: 'C'.