Q185 / 270 · NDA 2026 (Phase 1) · Mathematics
What is $\tan\left[2\tan^{-1}\dfrac{1}{2}-\dfrac{\pi}{4}\right]$ equal to ?
- A$-7$
- B$0$
- C$\dfrac{1}{5}$
- Correct answerD$\dfrac{1}{7}$
Solution
Let $\theta = \tan^{-1}\dfrac{1}{2}$, which implies $\tan\theta = \dfrac{1}{2}$.
We need to evaluate $\tan\left(2\theta - \dfrac{\pi}{4}\right)$.
First, calculate $\tan(2\theta)$ using the double angle formula:
$\tan(2\theta) = \dfrac{2\tan\theta}{1 - \tan^2\theta} = \dfrac{2\left(\dfrac{1}{2}\right)}{1 - \left(\dfrac{1}{2}\right)^2} = \dfrac{1}{1 - \dfrac{1}{4}} = \dfrac{4}{3}$
Now, apply the tangent subtraction formula:
$\tan\left(2\theta - \dfrac{\pi}{4}\right) = \dfrac{\tan(2\theta) - \tan\left(\dfrac{\pi}{4}\right)}{1 + \tan(2\theta)\tan\left(\dfrac{\pi}{4}\right)}$
Substitute $\tan(2\theta) = \dfrac{4}{3}$ and $\tan\left(\dfrac{\pi}{4}\right) = 1$:
$\tan\left(2\theta - \dfrac{\pi}{4}\right) = \dfrac{\dfrac{4}{3} - 1}{1 + \dfrac{4}{3}(1)} = \dfrac{\dfrac{1}{3}}{\dfrac{7}{3}} = \dfrac{1}{7}$
Answer: $\dfrac{1}{7}$
We need to evaluate $\tan\left(2\theta - \dfrac{\pi}{4}\right)$.
First, calculate $\tan(2\theta)$ using the double angle formula:
$\tan(2\theta) = \dfrac{2\tan\theta}{1 - \tan^2\theta} = \dfrac{2\left(\dfrac{1}{2}\right)}{1 - \left(\dfrac{1}{2}\right)^2} = \dfrac{1}{1 - \dfrac{1}{4}} = \dfrac{4}{3}$
Now, apply the tangent subtraction formula:
$\tan\left(2\theta - \dfrac{\pi}{4}\right) = \dfrac{\tan(2\theta) - \tan\left(\dfrac{\pi}{4}\right)}{1 + \tan(2\theta)\tan\left(\dfrac{\pi}{4}\right)}$
Substitute $\tan(2\theta) = \dfrac{4}{3}$ and $\tan\left(\dfrac{\pi}{4}\right) = 1$:
$\tan\left(2\theta - \dfrac{\pi}{4}\right) = \dfrac{\dfrac{4}{3} - 1}{1 + \dfrac{4}{3}(1)} = \dfrac{\dfrac{1}{3}}{\dfrac{7}{3}} = \dfrac{1}{7}$
Answer: $\dfrac{1}{7}$
Hence, the correct answer is option: 'D'.