Inverse Trigonometric Functions Important questions

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} $

(i) 11
(ii)$\frac {\pi}{3}$
(iii) 1
(iv) 0

$tan^{-1}( \frac { \sqrt {1+ cosx} + \sqrt {1-cos x}}{ \sqrt {1+ cosx} - \sqrt {1-cosx}}) = tan^{-1}( \frac { \sqrt {2cos^2 \frac {x}{2}} + \sqrt {2sin^2 \frac {x}{2}}}{ \sqrt {2cos^2 \frac {x}{2}} - \sqrt {2sin^2 \frac {x}{2}}})$
Now as $\pi < x < \frac {3\pi}{2}$
or $\pi/2 < x/2 < \frac {3\pi}{4}$
or $cos \frac {x}{2} < 0 $ and $sin \frac {x}{2} < 0 $
So above equation can be written as
$=tan^{-1}( \frac { -\sqrt 2 cos \frac {x}{2} + \sqrt 2 sin \frac {x}{2}}{ -\sqrt 2 cos \frac {x}{2} - \sqrt 2 sin \frac {x}{2}}$
$=tan^{-1}( \frac { cos \frac {x}{2} - sin \frac {x}{2}}{ cos \frac {x}{2} + sin \frac {x}{2}}$
$=tan^{-1}( \frac {1 - tan \frac {x}{2}}{1 + tan \frac {x}{2}}$
$=tan^{-1 ( tan (\frac {\pi}{4} - \frac {x}{2})) = \frac {\pi}{4} - \frac {x}{2}$

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
$2sin^{-1} \frac {3}{5} - tan^{-1} \frac {17}{31}= 2 tan^{-1} \frac {3}{4} - tan^{-1} \frac {17}{31}$
$=tan^{-1} \frac {2. \frac {3}{4}}{1 - \frac {9}{16}} - tan^{-1} \frac {17}{31}$
$=tan^{-1} \frac {24}{7} - tan^{-1} \frac {17}{31}$
$=tan^{-1} \frac { 24/7 - 17/31}{1+ (24/7)(17/31)} = tan ^{-1} 1 = \frac {\pi}{2}$

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
$x= \pm \frac {3}{4}$

$tan^{-1}(x-1) + tan^{-1} x = tan^{-1} 3x - tan^{-1} (x+1)$
$tan^{-1} \frac {x-1 +x}{1-x(x-1)} =tan^{-1} \frac {3x -(x+1)}{1+ 3x(1+x)}$
$tan^{-1} \frac {2x -1}{1-x^2+x}= tan^{-1} \frac {2x -1}{1+ 3x + 3x^2}$
or
$\frac {2x -1}{1-x^2+x}= \frac {2x -1}{1+ 3x + 3x^2}$
So 2x -1=0 or x=1/2
or
$1+ 3x + 3x^2 =1-x^2+x$
$4x^2 + 2x =0$
$2x^2 + x=0$
$x(2x+1)=0$
or x= 0,x=-1/2

$tan^{-1} (\frac {1-x/2}{1+x/2}) = \frac {1}{2} tan^{-1} \frac {x}{2}$
$tan^{-1} 1 - tan^{-1} \frac {x}{2}=\frac {1}{2} tan^{-1} \frac {x}{2}$
$\frac {\pi}{2} - 2 tan^{-1} \frac {x}{2}= tan^{-1} \frac {x}{2}$
$ tan^{-1} \frac {x}{2}= \frac {\pi}{6}$
$x =\frac {2}{\sqrt 3}$

$sin (sin^{-1} x + sin^{-1} (1 - x)) = sin (cos^{-1}x)$
$sin (sin^{-1} x) cos (sin^{-1} (1 – x)) + cos (sin^{-1} x) sin (sin^{-1} (1 - x) ) = sin (cos^{-1} x)$
$x\sqrt {1 -(1-x)^2} + \sqrt {1-x^2} (1-x) = \sqrt {1-x^2}$
x=0 or 1/2

Putting $x=cos 2 \theta$
$tan^{-1}( \frac { \sqrt {1+ cos 2 \theta } - \sqrt {1-cos 2 \theta}}{ \sqrt {1+ cos2 \theta} + \sqrt {1-cos2 \theta}}) = tan^{-1}( \frac { \sqrt {2cos^2 \theta} - \sqrt {2sin^2 \theta}}{ \sqrt {2cos^2 \theta} + \sqrt {2sin^2 \theta}})$
Now as $-\frac {1}{\sqrt 2} \leq x \leq 1$
or $-\frac {1}{\sqrt 2} \leq cos 2\theta \leq 1$
$ 0 \leq 2\theta \leq \frac {3\pi}{4}$
or $ 0 \leq \theta \leq \frac {3\pi}{8}$
So above equation can be written as
$=tan^{-1}( \frac { \sqrt 2 cos \theta - \sqrt 2 sin \theta}{ \sqrt 2 cos\theta + \sqrt 2 sin \theta}$
$=tan^{-1}( \frac {1 - tan \theta}{1 + tan \theta})$
$=tan^{-1} ( tan (\frac {\pi}{4} -\theta)) = \frac {\pi}{4} - \frac {1}{2} cos^{-1}x$

$tan^{-1} \frac {5x}{1-6x^2} = \frac {\pi}{4}$
$5x=1-6x^2$
$6x^2 -5x -1=0$
x=1 or x= 1/6
x=1/6 only satisfies this equation

​$sin ^{-1} \frac {4}{5} +sin ^{-1} \frac {5}{13} + sin ^{-1} \frac {16}{65}= tan ^{-1} \frac {4}{3} +tan ^{-1} \frac {5}{12} + tan ^{-1} \frac {16}{63}$
$=tan ^{-1} \frac {63}{16} + tan ^{-1} \frac {16}{63} = tan ^{-1} \frac {63}{16} + cot ^{-1} \frac {63}{16} = \frac {\pi}{2}$