Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\tan^{-1}\displaystyle\frac{2}{11} + \tan^{-1}\frac{7}{24}\) | (I) | \(\displaystyle\frac{3\pi}{4}\) |
| (B) | \(\tan^{-1}2 + \tan^{-1}3\) | (II) | \(\pi\) |
| (C) | \(\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3\) | (III) | \(\tan^{-1}\displaystyle\frac{1}{2}\) |
| (D) | \(\tan^{-1}\displaystyle\frac{1}{7} + \tan^{-1}\frac{1}{13}\) | (IV) | \(\tan^{-1}\displaystyle\frac{2}{9}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
For x, y > 0 the addition formula is used in two forms:
\( \displaystyle \tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x + y}{1 - xy} \quad (xy \lt 1) \)
\( \displaystyle \tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\frac{x + y}{1 - xy} \quad (xy \gt 1) \)
(A) \(\tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24}\)
This matches (III) \(\tan^{-1}\frac{1}{2}\). Here \(xy = \frac{14}{264} \lt 1\):
\( \displaystyle \frac{x + y}{1 - xy} = \frac{\frac{48 + 77}{264}}{\frac{264 - 14}{264}} = \frac{125}{250} = \frac{1}{2} \)
(B) \(\tan^{-1}2 + \tan^{-1}3\)
This matches (I) \(\frac{3\pi}{4}\). Here \(xy = 6 \gt 1\):
\( \displaystyle \tan^{-1}2 + \tan^{-1}3 = \pi + \tan^{-1}\frac{5}{1 - 6} \)
\( \displaystyle = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)
(C) \(\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3\)
This matches (II) \(\pi\). Here \(\tan^{-1}1 = \frac{\pi}{4}\) and \(\tan^{-1}2 + \tan^{-1}3 = \frac{3\pi}{4}\):
\( \displaystyle \frac{\pi}{4} + \frac{3\pi}{4} = \pi \)
(D) \(\tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{13}\)
This matches (IV) \(\tan^{-1}\frac{2}{9}\). Here \(xy = \frac{1}{91} \lt 1\):
\( \displaystyle \frac{x + y}{1 - xy} = \frac{\frac{20}{91}}{\frac{90}{91}} = \frac{2}{9} \)
So the matching is (A)-(III), (B)-(I), (C)-(II), (D)-(IV).
Tip: In (B), the formula without \(\pi\) gives \(\tan^{-1}(-1) = -\frac{\pi}{4}\), a negative value for a sum of two positive angles. When \(xy \gt 1\) (with x, y > 0), add \(\pi\).
Note: This topic was not in the CUET 2026 syllabus.