If \(2\tan^{-1}\displaystyle\frac{1}{2} + \tan^{-1}\displaystyle\frac{1}{7} = \tan^{-1}x\), then value of x is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{31}{17}\)
Two standard results are used:
\( \displaystyle 2\tan^{-1}a = \tan^{-1}\frac{2a}{1 - a^2}, \quad |a| \lt 1 \)
\( \displaystyle \tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a + b}{1 - ab}, \quad ab \lt 1 \)
With \(a = \frac{1}{2}\):
\( \displaystyle 2\tan^{-1}\frac{1}{2} = \tan^{-1}\frac{1}{1 - \frac{1}{4}} = \tan^{-1}\frac{4}{3} \)
Now \(\frac{4}{3} \cdot \frac{1}{7} = \frac{4}{21} \lt 1\), so the sum formula applies:
\( \displaystyle \tan^{-1}\frac{4}{3} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{21}} \)
\( \displaystyle = \tan^{-1}\frac{31/21}{17/21} = \tan^{-1}\frac{31}{17} \)
Hence \(x = \frac{31}{17}\).
Note: This topic was not in the CUET 2026 syllabus.