Online Test — Inverse Trigonometric Functions
10 Questions • 20 min • Chapter MCQ
20:00
Question 1 of 10
easy
The principal value of $\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ is:
$\frac{\pi}{3}$
$-\frac{\pi}{3}$
$\frac{2\pi}{3}$
$-\frac{2\pi}{3}$
Explanation: $\sin(-\pi/3)=-\sqrt{3}/2$ and $-\pi/3 \in [-\pi/2,\pi/2]$.
Question 2 of 10
easy
$\cos^{-1}(-x) + \cos^{-1}(x)$ equals:
$\pi$
$0$
$\pi/2$
$2\pi$
Explanation: $\cos^{-1}(-x)=\pi-\cos^{-1}x$, so $\cos^{-1}(-x)+\cos^{-1}(x)=\pi$.
Question 3 of 10
medium
$\sin^{-1}\left(\sin\frac{5\pi}{6}\right)$ equals:
$\frac{5\pi}{6}$
$\frac{\pi}{6}$
$-\frac{\pi}{6}$
$\frac{\pi}{3}$
Explanation: $\sin(5\pi/6)=\sin(\pi/6)=1/2$. So the answer is $\pi/6 \in [-\pi/2,\pi/2]$.
Question 4 of 10
easy
$\tan^{-1}\frac{1}{\sqrt{3}} + \cot^{-1}\frac{1}{\sqrt{3}}$ equals:
$\pi/6$
$\pi/3$
$\pi/2$
$\pi$
Explanation: By the property $\tan^{-1}x + \cot^{-1}x = \pi/2$ for all $x \in \mathbb{R}$.
Question 5 of 10
medium
The domain of $\sec^{-1}(2x)$ is:
$[-1/2,1/2]$
$|x| \ge 1/2$, i.e. $(-\infty,-1/2]\cup[1/2,\infty)$
$\mathbb{R}$
$[-1,1]$
Explanation: $\sec^{-1}$ requires $|\text{argument}|\ge1$, i.e. $|2x|\ge1$, so $|x|\ge1/2$.
Question 6 of 10
hard
$2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7}$ equals:
$\pi/4$
$\pi/3$
$\pi/2$
$\pi$
Explanation: $2\tan^{-1}(1/3)=\tan^{-1}\frac{2/3}{1-1/9}=\tan^{-1}\frac{2/3}{8/9}=\tan^{-1}\frac{3}{4}$. Then $\tan^{-1}(3/4)+\tan^{-1}(1/7)=\tan^{-1}\frac{3/4+1/7}{1-3/28}=\tan^{-1}\frac{25/28}{25/28}=\tan^{-1}(1)=\pi/4$.
Question 7 of 10
hard
If $\sin^{-1}x + \sin^{-1}y = \pi/2$, then $\cos^{-1}x + \cos^{-1}y$ equals:
$\pi$
$\pi/2$
$0$
$\pi/4$
Explanation: Since $\sin^{-1}x+\cos^{-1}x=\pi/2$ and $\sin^{-1}y+\cos^{-1}y=\pi/2$, adding: $(\sin^{-1}x+\sin^{-1}y)+(\cos^{-1}x+\cos^{-1}y)=\pi$. Given the first sum $=\pi/2$, the second $=\pi/2$.
Question 8 of 10
hard
The value of $\cos\left(\sin^{-1}\frac{3}{5}+\cos^{-1}\frac{4}{5}\right)$ is:
$\frac{24}{25}$
$\frac{7}{25}$
$0$
$1$
Explanation: Let $\alpha=\sin^{-1}(3/5)$, $\beta=\cos^{-1}(4/5)$. Then $\sin\alpha=3/5,\cos\alpha=4/5$; $\cos\beta=4/5,\sin\beta=3/5$. $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=16/25-9/25=7/25$. Wait, let me recalculate: $(4/5)(4/5)-(3/5)(3/5)=16/25-9/25=7/25$. Oh the answer is 7/25.
Question 9 of 10
hard
$\tan^{-1}(2) + \tan^{-1}(3)$ equals:
$\pi/4$
$3\pi/4$
$\pi/2$
$-\pi/4$
Explanation: Here $x=2,y=3$, $xy=6>1$ with $x,y>0$. So $\tan^{-1}2+\tan^{-1}3=\pi+\tan^{-1}\frac{2+3}{1-6}=\pi+\tan^{-1}(-1)=\pi-\pi/4=3\pi/4$.
Question 10 of 10
medium
The value of $\sin^{-1}(\cos(\sin^{-1}(\sqrt{3}/2)))$ is:
$\pi/3$
$\pi/6$
$\pi/4$
$\pi/2$
Explanation: $\sin^{-1}(\sqrt{3}/2)=\pi/3$. $\cos(\pi/3)=1/2$. $\sin^{-1}(1/2)=\pi/6$.