Inverse Trigonometric Functions Important questions
Multiple Choice Questions
Question 1
What is the range of the inverse sine function $sin^{-1} x$
(a) $-\pi \leq sin^{-1} x \leq \pi$
(b) $-\frac {\pi}{2} \leq sin^{-1} x \leq \frac {\pi}{2} $
(c) $0 \leq sin^{-1} x \leq \leq \pi $
(d) $-\pi \ge sin^{-1} x < \pi$
Question 2
Which of the following statements is true about the arc cosine function?
(a) $cos^{-1}(-x) = -cos^{-1}x $
(b) $cos^{-1}(-x) = \pi - cos^{-1}x $
(c) $cos^{-1}(-x) = cos^{-1}x $
(d) $cos^{-1}(-x) = \pi + cos^{-1}x$
Question 3
If x and y are positive acute angles such that cos(x) = sin(y), then which of the following is true?
(a) $x + y = \pi/2$
(b) x = y
(c) $x = \pi/2 - y$
(d) $x + y = \pi$
Question 4
What is the value of $sin^{-1}(-1/2)$?
(a) -π/6
(b) π/6
(c) -π/3
(d) π/3
Question 5
If $tan^{-1}(x) = \theta$, then which of the following is true?
(a) $sin(\theta) = \frac {x}{\sqrt{(1 + x^2)}}$
(b) $sin(\theta) = \frac {x}{\sqrt {(1 - x^2)}}$
(c) $cos(\theta) = \frac {x}{\sqrt {(1 + x^2)}}$
(d) $cos(\theta) = \frac {x}{\sqrt {(1 - x^2)}}$
Question 6
The domain of the function $y = sin^{–1} (-x^2)$ is
(a)[0, 1]
(b) (0, 1)
(c) [–1, 1]
(d) $\phi$
Question 7
If $sin^{-1} x + sin^{-1} y = \frac {\pi}{2}$ , then value of $cos^{-1} x + cos^{-1} y$ is
(a) $\frac {\pi}{2}$
(b) $-\frac {\pi}{2}$
(d) 0
(d) $\pi$
Question 8
if $ A= tan^{-1} tan \frac {5\pi}{4}$ and $B= tan^{-1} (-tan \frac {2\pi}{3})$, then
(a) 2A=3B
(b) 4A=3B
(c) A=2B
(d) 6A=B
Question 9
The greatest and least value of the function $f(x) = (sin^{-1}x)^2 + (cos^{-1}x)^2$ are
(a)$\frac{\pi ^2}{4}$, 0
(b) $\frac{5\pi ^2}{4}$, $\frac{\pi ^2}{8}$
(c) $\frac{\pi ^2}{4}$, $\frac{\pi ^2}{8}$
(d) $\frac{3\pi ^2}{4}$, $\frac{\pi ^2}{8}$
Question 10
The value of $sin^{-1} ( cos \frac {33\pi}{5})$ is
(a) $\frac {3\pi}{5}$
(b) $-\frac {\pi}{10}$
(c) $\frac {\pi}{10}$
(d) $\frac {7\pi}{5}$ Answers 1-10
1. b
2. b
3. a
4. a
5. a
6. c
7. a
8. b
9. b
$f(x) = (sin^{-1}x)^2 + (cos^{-1}x)^2 = (sin^{-1} x + cos^{-1} x) ^2 - 2 sin^{-1} x cos^{-1} x$
$= \frac {\pi^2}{4} - 2 sin^{-1} x ( \frac {\pi}{2} - sin^{-1} x)= \frac {\pi^2}{4} - \pi sin^{-1} x +2 (sin^{-1} x)^2$
$= 2((sin^{-1} x)^2 - \frac {\pi }{2} sin^{-1} x \frac {\pi^2}{8})= 2[(sin^{-1} x - \frac {\pi}{4})^2 + \frac {\pi^2}{16}]$
So greatest and least value of the function are $\frac{5\pi ^2}{4}$, $\frac{\pi ^2}{8}$
10. b
$sin^{-1} ( cos \frac {33\pi}{5})= sin^{-1} [ cos ({6\pi + \frac {3\pi}{5}]= sin^{-1} cos( \frac {3\pi}{5}) = sin^{-1} sin(\frac {\pi}{2}- \frac {3\pi}{5})$
$=sin^{-1} sin(\frac {-\pi}{10})=\frac {-\pi}{10} $
Fill in the blanks
Question 11
(i) The value of $tan^2(sec^{–1} 2) + cot^2 (cosec^{–1} 3)$ is ________
(ii) The principal value of $tan^{-1} \sqrt 3$ is _____
(iii) The value of $sin (sin^{-1} x + cos^{-1} x)$ , $|x| \leq 1$ is ________
(iv) The value of $cos^{-1} (cos \frac {5\pi}{3}) + sin^{-1} (sin \frac {5\pi}{3})$ is _____ Answers
Here we will convert $sin^{-1}$ into $tan^{-1}$
Let $sin^{-1} \frac {3}{5}=x$
$sin x = \frac {3}{5}$ here $x \in [-\frac {\pi}{2},\frac {\pi}{2}]$
So $tan x = \frac {sin x}{\sqrt {1 - sin^2 x}} = \frac {3}{4}$
So $x = tan^{-1} \frac {3}{4}$
So LHS becomes
$2sin^{-1} \frac {3}{5} - tan^{-1} \frac {17}{31}= 2 tan^{-1} \frac {3}{4} - tan^{-1} \frac {17}{31}$
$=tan^{-1} \frac {2. \frac {3}{4}}{1 - \frac {9}{16}} - tan^{-1} \frac {17}{31}$
$=tan^{-1} \frac {24}{7} - tan^{-1} \frac {17}{31}$
$=tan^{-1} \frac { 24/7 - 17/31}{1+ (24/7)(17/31)} = tan ^{-1} 1 = \frac {\pi}{2}$
Question 14
Solve the equation
$cos(tan^{-1} x) = sin (cot^{-1} \frac {3}{4})$ Solution
Let $tan^{-1} x=y$ or $tan y=x$
or $cos y=\frac {1}{\sqrt {x^2 + 1}}$
or $y = cos^{-1} \frac {1}{\sqrt {x^2 + 1}}$
Let $cot^{-1} \frac {3}{4}=p$
or $cot p = 3/4$
or $sin p = 4/5$
or $p=sin^{-1} 4/5$
So the above equation can be written as
$\frac {1}{\sqrt {x^2 + 1}}= \frac {4}{5}$
Squaring both side, we get
$x= \pm \frac {3}{4}$
Question 15
Solve the equation
$tan^{-1} (x-1) + tan^{-1} x + tan^{-1} (x+1) =tan^{-1} 3x$ Solution
