Match List - I with List - II, List - I, List - II, A. ^(-1)((7)/6), I. (5)/6, B. ^(-1)((8)/6), II. ()/2, C. ^(-1)1/√3 + cosec^(-1)2/√3, III. ()/6, D. ^(-1)((5)/6), IV. ()/3, Choose the correct answer from the options given below: Option 1: A-III, B-II, C-IV, D-I Option 2: A-I, B-II, C-IV, D-III Option 3: A-III, B-IV, C-II, D-I Option 4: A-I, B-II, C-III, D-IV

Answer: A-III, B-IV, C-II, D-I. Solution: A: (7/6) = (/6) = 1/√3, so ^(-1)(1/√3) = /6 - III. B: (8/6) = (4/3) = (/3) = √3, so result = /3 - IV. C: /6 + /3 = /2 - II. D: 5/6 [0,], so result = 5/6 - I. So A-III, B-IV, C-II, D-I.

Geometry

Trigonometry

Medium

Match List - I with List - II

List - I List - II
A. $\tan^{-1}\left(\tan\frac{7\pi}{6}\right)$ I. $\frac{5\pi}{6}$
B. $\tan^{-1}\left(\tan\frac{8\pi}{6}\right)$ II. $\frac{\pi}{2}$
C. $\tan^{-1}\frac{1}{\sqrt{3}} + cosec^{-1}\frac{2}{\sqrt{3}}$ III. $\frac{\pi}{6}$
D. $\cos^{-1}\left(\cos\frac{5\pi}{6}\right)$ IV. $\frac{\pi}{3}$

Choose the correct answer from the options given below:

List - I	List - II
A. $\tan^{-1}\left(\tan\frac{7\pi}{6}\right)$	I. $\frac{5\pi}{6}$
B. $\tan^{-1}\left(\tan\frac{8\pi}{6}\right)$	II. $\frac{\pi}{2}$
C. $\tan^{-1}\frac{1}{\sqrt{3}} + cosec^{-1}\frac{2}{\sqrt{3}}$	III. $\frac{\pi}{6}$
D. $\cos^{-1}\left(\cos\frac{5\pi}{6}\right)$	IV. $\frac{\pi}{3}$

A-III, B-II, C-IV, D-I

A-I, B-II, C-IV, D-III

A-III, B-IV, C-II, D-I

A-I, B-II, C-III, D-IV

✅ Correct Option: 3

A: $\tan(7\pi/6) = \tan(\pi/6) = 1/\sqrt{3}$ , so $\tan^{-1}(1/\sqrt{3}) = \pi/6$ - III.

B: $\tan(8\pi/6) = \tan(4\pi/3) = \tan(\pi/3) = \sqrt{3}$ , so result $= \pi/3$ - IV.

C: $\pi/6 + \pi/3 = \pi/2$ - II.

D: $5\pi/6 \in [0,\pi]$ , so result $= 5\pi/6$ - I.

So A-III, B-IV, C-II, D-I.

2023: 30 May Shift 3

If $\sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3}$ , then the value of $\cos^{-1} x + \cos^{-1} y$ is :

2023: 23 May Shift 3

The principal value of $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

2022: 10 Aug Shift 1

The value of $\frac{1}{AB^2} + \frac{1}{BC^2}$ is:

2024: 16 May Shift 1

If $\tan ^{-1}\left(\frac{2}{3^{-x}+1}\right)=\cot ^{-1}\left(\frac{3}{3^{x}+1}\right)$ , then which one of the following is true ?

2026: 18 May Shift 2

Match the LIST-I with LIST-II

LIST-I Functions	LIST-II Principal value
A. $\sin^{-1}\left(\dfrac{1}{2}\right)$	I. $\dfrac{\pi}{4}$
B. $\sin^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)$	II. $\dfrac{\pi}{6}$
C. $\tan^{-1}(1)$	III. $-\dfrac{\pi}{4}$
D. $\cos^{-1}\left(-\dfrac{1}{2}\right)$	IV. $\dfrac{2\pi}{3}$

Choose the correct answer from the options given below:

2026: 22 May Shift 2

$\cos[\tan^{-1}\{\sin(\cot^{-1}x)\}]$ is equal to

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)