Q177 / 270 · NDA 2026 (Phase 1) · Mathematics
If $\alpha$ and $\beta$ are complementary angles such that $\alpha-\beta=\dfrac{\pi}{6}$ and $m\tan\beta=n\tan\alpha$, then what is $\left(\dfrac{m+n}{m-n}\right)$ equal to ?
- Correct answerA$2$
- B$\dfrac{2}{\sqrt{3}}$
- C$1$
- D$\dfrac{1}{\sqrt{3}}$
Solution
Since $\alpha$ and $\beta$ are complementary angles, $\alpha + \beta = \dfrac{\pi}{2}$.
We are given $\alpha - \beta = \dfrac{\pi}{6}$.
Adding the two equations yields $2\alpha = \dfrac{2\pi}{3} \Rightarrow \alpha = \dfrac{\pi}{3}$.
Subtracting the two equations yields $2\beta = \dfrac{\pi}{3} \Rightarrow \beta = \dfrac{\pi}{6}$.
From the given relation $m\tan\beta = n\tan\alpha$, we have:
$\dfrac{m}{n} = \dfrac{\tan\alpha}{\tan\beta} = \dfrac{\tan\left(\dfrac{\pi}{3}\right)}{\tan\left(\dfrac{\pi}{6}\right)} = \dfrac{\sqrt{3}}{\dfrac{1}{\sqrt{3}}} = 3$
Therefore, $\dfrac{m+n}{m-n} = \dfrac{3+1}{3-1} = 2$.
Answer: $2$
We are given $\alpha - \beta = \dfrac{\pi}{6}$.
Adding the two equations yields $2\alpha = \dfrac{2\pi}{3} \Rightarrow \alpha = \dfrac{\pi}{3}$.
Subtracting the two equations yields $2\beta = \dfrac{\pi}{3} \Rightarrow \beta = \dfrac{\pi}{6}$.
From the given relation $m\tan\beta = n\tan\alpha$, we have:
$\dfrac{m}{n} = \dfrac{\tan\alpha}{\tan\beta} = \dfrac{\tan\left(\dfrac{\pi}{3}\right)}{\tan\left(\dfrac{\pi}{6}\right)} = \dfrac{\sqrt{3}}{\dfrac{1}{\sqrt{3}}} = 3$
Therefore, $\dfrac{m+n}{m-n} = \dfrac{3+1}{3-1} = 2$.
Answer: $2$
Hence, the correct answer is option: 'A'.