12th NCERT Inverse Trigonometric Functions Miscellaneous Exercise Questions 17
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Find the value of the following:

Question (1)

${\cos ^{ - 1}}\left( {\cos \frac{{13\pi }}{6}} \right)$

Solution

${\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta ,\theta \in \left[ {0,\pi } \right]$
${\cos ^{ - 1}}\left( {\cos \frac{{13\pi }}{6}} \right)$ $= {\cos ^{ - 1}}\left[ {\cos \left( {2\pi + \frac{\pi }{6}} \right)} \right]$ $= {\cos ^{ - 1}}\left( {\cos \frac{\pi }{6}} \right)$ $= \frac{\pi }{6}$

Question (2)

${\tan ^{ - 1}}\left[ {\tan \frac{{7\pi }}{6}} \right]$

Solution

${\tan ^{ - 1}}\left[ {\tan \theta } \right] = \theta ,\quad \theta \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$
${\tan ^{ - 1}}\left[ {\tan \frac{{7\pi }}{6}} \right]$ $= {\tan ^{ - 1}}\left[ {\tan \left( {\pi + \frac{\pi }{6}} \right)} \right]$ $= {\tan ^{ - 1}}\left[ {\tan \frac{\pi }{6}} \right] = \frac{\pi }{6}$

### Prove that

Question (3)

$2{\sin ^{ - 1}}\frac{3}{5} = {\tan ^{ - 1}}\frac{{24}}{7}$

Solution

${\sin ^{ - 1}}x = {\tan ^{ - 1}}\frac{x}{{\sqrt {1 - {x^2}} }}$
$LHS = 2{\sin ^{ - 1}}\frac{3}{5}$ $= 2{\tan ^{ - 1}}\frac{3}{4}$ $= {\tan ^{ - 1}}\frac{3}{4} + {\tan ^{ - 1}}\frac{3}{4}$ $= {\tan ^{ - 1}}\left[ {\frac{{\frac{3}{4} + \frac{3}{4}}}{{1 - \frac{3}{4} \times \frac{3}{4}}}} \right]$ $= {\tan ^{ - 1}}\left( {\frac{{\frac{6}{4}}}{{\frac{{16 - 9}}{{16}}}}} \right)$ $= {\tan ^{ - 1}}\frac{{24}}{7} = RHS$

Question (4)

${\sin ^{ - 1}}\frac{8}{{17}} + {\sin ^{ - 1}}\frac{3}{5} = {\tan ^{ - 1}}\frac{{77}}{{36}}$

Solution

${\sin ^{ - 1}}x = {\tan ^{ - 1}}\frac{x}{{\sqrt {1 - {x^2}} }}$
$LHS = {\sin ^{ - 1}}\frac{8}{{17}} + {\sin ^{ - 1}}\frac{3}{5}$ $= {\tan ^{ - 1}}\frac{8}{{15}} + {\tan ^{ - 1}}\frac{3}{4}$ $= {\tan ^{ - 1}}\left[ {\frac{{\frac{8}{{15}} + \frac{3}{4}}}{{1 - \frac{8}{{15}} \times \frac{3}{4}}}} \right]$ $= {\tan ^{ - 1}}\left( {\frac{{32 + 45}}{{60 - 24}}} \right)$ $= {\tan ^{ - 1}}\frac{{77}}{{36}} = RHS$

Question (5)

${\cos ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{{12}}{{13}} = {\cos ^{ - 1}}\frac{{33}}{{65}}$

Solution

${\cos ^{ - 1}}x = {\tan ^{ - 1}}\frac{{\sqrt {1 - {x^{}}} }}{x},{\tan ^{ - 1}}x = {\cos ^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}$
$LHS = {\cos ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{{12}}{{13}}$ $= {\tan ^{ - 1}}\frac{3}{4} + {\tan ^{ - 1}}\frac{5}{{12}}$ $= {\tan ^{ - 1}}\left[ {\frac{{\frac{3}{4} + \frac{5}{{12}}}}{{1 - \frac{3}{4} \times \frac{5}{{12}}}}} \right]$ $= {\tan ^{ - 1}}\left( {\frac{{36 + 20}}{{48 - 15}}} \right)$ $= {\tan ^{ - 1}}\frac{{56}}{{33}}$ $= {\cos ^{ - 1}}\frac{{33}}{{65}} = RHS$

Question (6)

${\cos ^{ - 1}}\frac{{12}}{{13}} + {\sin ^{ - 1}}\frac{3}{5} = {\sin ^{ - 1}}\frac{{56}}{{65}}$

Solution

${\cos ^{ - 1}}x = {\tan ^{ - 1}}\frac{{\sqrt {1 - {x^2}} }}{x},{\tan ^{ - 1}}x = {\sin ^{ - 1}}\frac{x}{{\sqrt {1 + {x^2}} }},{\sin ^{ - 1}}x = {\tan ^{ - 1}}\frac{x}{{\sqrt {1 - {x^2}} }}$
$LHS = {\cos ^{ - 1}}\frac{{12}}{{13}} + {\sin ^{ - 1}}\frac{3}{5}$ $= {\tan ^{ - 1}}\frac{5}{{12}} + {\tan ^{ - 1}}\frac{3}{4}$ $= {\tan ^{ - 1}}\left[ {\frac{{\frac{5}{{12}} + \frac{3}{4}}}{{1 - \frac{5}{{12}} \times \frac{3}{4}}}} \right]$ $= {\tan ^{ - 1}}\left( {\frac{{20 + 36}}{{48 - 15}}} \right)$ $= {\tan ^{ - 1}}\frac{{56}}{{33}}$ $= {\sin ^{ - 1}}\frac{{56}}{{65}} = RHS$

Question (7)

${\tan ^{ - 1}}\frac{{63}}{{16}} = {\sin ^{ - 1}}\frac{5}{{13}} + {\cos ^{ - 1}}\frac{3}{5}$

Solution

${\cos ^{ - 1}}x = {\tan ^{ - 1}}\frac{{\sqrt {1 - {x^2}} }}{x},,{\sin ^{ - 1}}x = {\tan ^{ - 1}}\frac{x}{{\sqrt {1 - {x^2}} }}$
$RHS = {\sin ^{ - 1}}\frac{5}{{13}} + {\cos ^{ - 1}}\frac{3}{5}$ $= {\tan ^{ - 1}}\frac{5}{{12}} + {\tan ^{ - 1}}\frac{4}{3}$ $= {\tan ^{ - 1}}\left[ {\frac{{\frac{5}{{12}} + \frac{4}{3}}}{{1 - \frac{5}{{12}} \times \frac{4}{3}}}} \right]$ $= {\tan ^{ - 1}}\left( {\frac{{15 + 48}}{{36 - 20}}} \right)$ $= {\tan ^{ - 1}}\frac{{63}}{{16}} = LHS$

Question (8)

${\tan ^{ - 1}}\frac{1}{5} + {\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{8} = \frac{\pi }{4}$

Solution

${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)$
$LHS = {\tan ^{ - 1}}\frac{1}{5} + {\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{8}$ $= {\tan ^{ - 1}}\left[ {\frac{{\frac{1}{5} + \frac{1}{7}}}{{1 - \frac{1}{5} \times \frac{1}{7}}}} \right] + {\tan ^{ - 1}}\left[ {\frac{{\frac{1}{3} + \frac{1}{8}}}{{1 - \frac{1}{3} \times \frac{1}{8}}}} \right]$ $= {\tan ^{ - 1}}\left( {\frac{{7 + 5}}{{35 - 1}}} \right) + {\tan ^{ - 1}}\left( {\frac{{8 + 3}}{{24 - 1}}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{12}}{{34}}} \right) + {\tan ^{ - 1}}\left( {\frac{{11}}{{23}}} \right)$ $= {\tan ^{ - 1}}\left[ {\frac{{\frac{{12}}{{34}} + \frac{{11}}{{23}}}}{{1 - \frac{{12}}{{34}} \times \frac{{11}}{{23}}}}} \right]$ $= {\tan ^{ - 1}}\left( {\frac{{276 + 374}}{{782 - 132}}} \right)$ $= {\tan ^{ - 1}}\frac{{650}}{{650}}$ $= {\tan ^{ - 1}}1$ $= \frac{\pi }{4} = RHS$ Prove that

Question (9)

${\tan ^{ - 1}}\sqrt x = \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right),x \in \left[ {0,1} \right]$

Solution

$\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos (2\theta )$
$Let\quad x = {\tan ^2}\theta \; \Rightarrow \tan \theta = \sqrt x ,\quad \theta = {\tan ^{ - 1}}\sqrt x$ $RHS = \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right)$ $= \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$ $= \frac{1}{2}{\cos ^{ - 1}}\left( {\cos 2\theta } \right)$ $= \frac{1}{2}(2\theta ) = \theta$ $= {\tan ^{ - 1}}\sqrt x = LHS$

Question (10)

${\cot ^{ - 1}}\left( {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right) = \frac{x}{2},x \in \left( {0,\frac{\pi }{4}} \right)$

Solution

$x \in \left( {0,\frac{\pi }{4}} \right)\sqrt {1 + \sin x} = \cos \left( {\frac{x}{2}} \right) + \sin \left( {\frac{x}{2}} \right),\sqrt {1 - \sin x} = \cos \left( {\frac{x}{2}} \right) - \sin \left( {\frac{x}{2}} \right)$
${\cot ^{ - 1}}\left( {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right)$ $= {\cot ^{ - 1}}\left( {\frac{{\cos \left( {\frac{x}{2}} \right) + \sin \left( {\frac{x}{2}} \right) + \cos \left( {\frac{x}{2}} \right) - \sin \left( {\frac{x}{2}} \right)}}{{\cos \left( {\frac{x}{2}} \right) + \sin \left( {\frac{x}{2}} \right) - \left( {\cos \left( {\frac{x}{2}} \right) - \sin \left( {\frac{x}{2}} \right)} \right)}}} \right)$ $= {\cot ^{ - 1}}\left( {\frac{{2\cos \left( {\frac{x}{2}} \right)}}{{2\sin \left( {\frac{x}{2}} \right)}}} \right)$ $= {\cot ^{ - 1}}\left( {\cot \frac{x}{2}} \right)$ $= \frac{x}{2} = RHS$

Question (11)

${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {1 + x} + \sqrt {1 - x} }}} \right) = \frac{\pi }{4} - \frac{1}{2}{\cos ^{ - 1}}x, - \frac{1}{{\sqrt 2 }} \le x \le 1$

Solution

$1 + \cos 2\theta = 2{\cos ^2}\theta ,1 - \cos 2\theta = 2{\sin ^2}\theta$ ${\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}x - {\tan ^{ - 1}}y$
$Let\;x = \cos 2\theta \; \Rightarrow 2\theta = {\cos ^{ - 1}}x,\theta = \frac{1}{2}{\cos ^{ - 1}}x$ $- \frac{1}{{\sqrt 2 }} \le x \le 1,$ $- \frac{1}{{\sqrt 2 }} \le \cos 2\theta \le 1$ ${\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right) \le 2\theta \le {\cos ^{ - 1}}1$ $2\theta \in \left( {0,\frac{{3\pi }}{4}} \right)$ $\theta \in \left( {0,\frac{{3\pi }}{8}} \right)$ $LHS = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {1 + x} + \sqrt {1 - x} }}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + \cos 2\theta } - \sqrt {1 - \cos 2\theta } }}{{\sqrt {1 + \cos 2\theta } + \sqrt {1 - \cos 2\theta } }}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{\sqrt {2{{\cos }^2}\theta } - \sqrt {2{{\sin }^2}\theta } }}{{\sqrt {2{{\cos }^2}\theta } + \sqrt {2{{\sin }^2}\theta } }}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{\sqrt 2 \left( {\cos \theta - \sin \theta } \right)}}{{\sqrt 2 \left( {\cos \theta + \sin \theta } \right)}}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right)$ $= {\tan ^{ - 1}}1 - {\tan ^{ - 1}}\left( {\tan \theta } \right)$ $= \frac{\pi }{4} - \theta$ $= \frac{\pi }{4} - \frac{1}{2}{\cos ^{ - 1}}x$

Question (12)

$\frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3} = \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}$

Solution

${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2} \Rightarrow \frac{\pi }{2} - {\sin ^{ - 1}}x = {\cos ^{ - 1}}x$ ${\cos ^{ - 1}}x = {\sin ^{ - 1}}\sqrt {1 - {x^2}}$
$LHS = \frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3}$ $= \frac{9}{4}\left[ {\frac{\pi }{2} - {{\sin }^{ - 1}}\frac{1}{3}} \right]$ $= \frac{9}{4}{\cos ^{ - 1}}\frac{1}{3}$ $= \frac{9}{4}{\sin ^{ - 1}}\sqrt {1 - \frac{1}{9}}$ $= \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3} = RHS$ Solve the following equations.

Question (13)

$2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2\cos ecx} \right)$

Solution

$2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2\cos ecx} \right)$ ${\tan ^{ - 1}}\left( {\cos x} \right) + {\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2\cos ecx} \right)$ ${\tan ^{ - 1}}\left( {\frac{{\cos x + \cos x}}{{1 - {{\cos }^2}x}}} \right) = {\tan ^{ - 1}}\left( {2\cos ecx} \right)$ ${\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{{{\sin }^2}x}}} \right) = {\tan ^{ - 1}}\left( {\frac{2}{{\sin x}}} \right)$ $\left( {\frac{{2\cos x}}{{{{\sin }^2}x}}} \right) = \frac{2}{{\sin x}}$ $\begin{array}{l}\cot x = 1\\\tan x = 1\end{array}$ $x = {\tan ^{ - 1}}1 = \frac{\pi }{4}$

Question (14)

${\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x(x > 0)$

Solution

${\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x$ $2{\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = {\tan ^{ - 1}}x$ ${\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} + {\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = {\tan ^{ - 1}}x$ ${\tan ^{ - 1}}\left[ {\frac{{\frac{{2 - 2x}}{{1 + x}}}}{{1 - \frac{{{{(1 - x)}^2}}}{{{{(1 + x)}^2}}}}}} \right] = {\tan ^{ - 1}}x$ $\frac{{2 - 2x}}{{1 + x}} \times \frac{{{{\left( {1 + x} \right)}^2}}}{{1 + 2x + {x^2} - 1 + 2x - {x^2}}} = x$ $\frac{{2(1 - x)(1 + x)}}{{4x}} = x$ $\begin{array}{l}1 - {x^2} = 2{x^2}\\3{x^2} = 1\end{array}$ ${x^2} = \frac{1}{3} \Rightarrow x = \frac{1}{{\sqrt 3 }}$

Question (15)

$\sin ({\tan ^{ - 1}}x),|x| < 1\;is\;equal\;to$ (A) $\frac{x}{{\sqrt {1 - {x^2}} }}$   (B) $\frac{1}{{\sqrt {1 - {x^2}} }}$   (C) $\frac{1}{{\sqrt {1 + {x^2}} }}$ (D) $\frac{x}{{\sqrt {1 + {x^2}} }}$

Solution

$\sin ({\tan ^{ - 1}}x)$ $= \sin \left( {{{\sin }^{ - 1}}\frac{x}{{\sqrt {1 + {x^2}} }}} \right)$ $= \frac{x}{{\sqrt {1 + {x^2}} }}$ So (D) is the correct option.

Question (16)

${\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \frac{\pi }{2},then\;x\;is\;equal\;to$ (A) 0, 1/2   (B) 1,1/2   (C) 0   (D) 1/2

Solution

${\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}$ ${\sin ^{ - 1}}(1 - x) = \frac{\pi }{2} + 2{\sin ^{ - 1}}x$ $\sin \left( {{{\sin }^{ - 1}}(1 - x)} \right) = \sin \left( {\frac{\pi }{2} + 2{{\sin }^{ - 1}}x} \right)$ $1 - x = \cos (2{\sin ^{ - 1}}x)$ $Let\;{\sin ^{ - 1}}x = \theta \; \Rightarrow x = \sin \theta$ $1 - \sin \theta = \cos 2\theta$ $1 - \sin \theta = 1 - 2{\sin ^2}\theta$ $2{\sin ^2}\theta - \sin \theta = 0$ $\sin \theta (2\sin \theta - 1) = 0$ $\sin \theta = 0,\;\;\sin \theta = 1/2$ $x = 0,x = 1/2.$ But x = 1/2 does not satisfy the equation . So it is not possible .
So x = 0.
So C is the correct option.

Question (17)

${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - {\tan ^{ - 1}}\left( {\frac{{x - y}}{{x + y}}} \right)\;is\;equal\;to\;$ (A)$\frac{\pi }{2}$   (B) $\frac{\pi }{3}$   (C) $\frac{\pi }{4}$   (D) $\frac{-3\pi }{4}$

Solution

${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - {\tan ^{ - 1}}\left( {\frac{{x - y}}{{x + y}}} \right)$ $= {\tan ^{ - 1}}\left[ {\frac{{\frac{x}{y} - \frac{{x - y}}{{x + y}}}}{{1 + \frac{x}{y}\left( {\frac{{x - y}}{{x + y}}} \right)}}} \right]$ $= {\tan ^{ - 1}}\left[ {\frac{{x(x + y) - y(x - y)}}{{y(x + y) + x(x - y)}}} \right]$ $= {\tan ^{ - 1}}\left[ {\frac{{{x^2} + xy - yx - {y^2}}}{{yx + {y^2} + {x^2} - xy}}} \right]$ $= {\tan ^{ - 1}}\left[ {\frac{{{x^2} + {y^2}}}{{{x^2} + y}}} \right]$ $= {\tan ^{ - 1}}(1) = \frac{\pi }{4}$ So C is the correct answer.