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.
Exercise 2.2←
⇒ Miscellaneous Exercise