Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

Match List I with List II

List I List II
A. $\sin^{-1}x+]cos^{-1}x,x∈[-1,1]$ I. $-\frac{π}{2}$
B. $\tan^{-1}\sqrt{3}-\cot^{-1}(\sqrt{3})$ II. $-\frac{π}{6}$
C. $\cos^{-1}\left(\cos\frac{13π}{6}\right)$ III. $\frac{π}{2}$
D. $\sin^{-1}(-\frac{1}{2})$ IV. $\frac{π}{6}$

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

(A) $\sin^{-1}x+]cos^{-1}x=\frac{π}{2}$

(B) B. $\tan^{-1}\sqrt{3}-\cot^{-1}(\sqrt{3})$

$\frac{π}{3}-π+\cot^{-1}\sqrt{3}$

$=\frac{π}{3}-π-\frac{π}{6}=-\frac{π}{2}$

(C) $\cos^{-1}\left(\cos\frac{13π}{6}\right)=\cos^{-1}\left(\cos\left(2π+\frac{π}{6}\right)\right)$

$=\cos^{-1}\cos\frac{π}{6}=\frac{π}{6}$

(D) $\sin^{-1}(-\frac{1}{2})=-\frac{π}{6}$