12th NCERT Inverse Trigonometric Functions Exercise 2.2 Questions 21
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Question (1)

$3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right),x \in \left[ { - \frac{1}{2},\frac{1}{2}} \right]$

Solution

$\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta$
$Let\;{\sin ^{ - 1}}x = \theta \quad \Rightarrow x = \sin \theta$ $x \in \left[ { - \frac{1}{2},\frac{1}{2}} \right] \Rightarrow Sin\theta = \left[ { - \frac{1}{2},\frac{1}{2}} \right]$ $\theta \in \left[ { - \frac{\pi }{6},\frac{\pi }{6}} \right] \Rightarrow 3\theta \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$ $RHS = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)$ $= {\sin ^{ - 1}}\left( {3\sin \theta - 4{{\sin }^3}\theta } \right)$ $= {\sin ^{ - 1}}\left( {\sin 3\theta } \right)$ $= 3\theta = 3{\sin ^{ - 1}}x = LHS$ So $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right),x \in \left[ { - \frac{1}{2},\frac{1}{2}} \right]$

Question (2)

$3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)$

Solution

$\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta$
$Let\;{\cos ^{ - 1}}x = \theta \quad \Rightarrow x = \cos \theta$ $x \in \left[ {\frac{1}{2},1} \right] \Rightarrow \cos \theta = \left[ {\frac{1}{2},1} \right]$ $\theta \in \left[ {0,\frac{\pi }{3}} \right] \Rightarrow 3\theta \in \left[ {0,\pi } \right]$ $RHS = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)$ $= {\cos ^{ - 1}}\left( {4{{\cos }^3}\theta - 3\cos \theta } \right)$ $= {\cos ^{ - 1}}\left( {\cos 3\theta } \right)$ $= 3\theta = 3{\cos ^{ - 1}}x = LHS$ $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)$

Question (3)

${\tan ^{ - 1}}\left( {\frac{2}{{11}}} \right) + {\tan ^{ - 1}}\left( {\frac{7}{{24}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$

Solution

${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)$
$LHS = {\tan ^{ - 1}}\left( {\frac{2}{{11}}} \right) + {\tan ^{ - 1}}\left( {\frac{7}{{24}}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{\frac{2}{{11}} + \frac{7}{{24}}}}{{1 - \left( {\frac{2}{{11}}} \right)\left( {\frac{7}{{24}}} \right)}}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{\frac{{48 + 77}}{{11(24)}}}}{{\frac{{264 - 14}}{{11(24)}}}}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{125}}{{250}}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{1}{2}} \right) = RHS$ ${\tan ^{ - 1}}\left( {\frac{2}{{11}}} \right) + {\tan ^{ - 1}}\left( {\frac{7}{{24}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$

Question (4)

$2{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = {\tan ^{ - 1}}\left( {\frac{{31}}{{17}}} \right)$

Solution

${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)$
$LHS = 2{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{\frac{1}{2} + \frac{1}{2}}}{{1 - \left( {\frac{1}{2}} \right)\left( {\frac{1}{2}} \right)}}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{4}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{\frac{4}{3} + \frac{1}{7}}}{{1 - \left( {\frac{4}{3}} \right)\left( {\frac{1}{7}} \right)}}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{\frac{{28 + 3}}{{21}}}}{{\frac{{21 - 4}}{{21}}}}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{31}}{{17}}} \right) = RHS$ $2{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = {\tan ^{ - 1}}\left( {\frac{{31}}{{17}}} \right)$ Write the following functions in the simplest form:

Question (5)

${\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x}$

Solution

The trigonometric formulas to be used $1 - \cos \theta = 2{\sin ^2}\left( {\frac{\theta }{2}} \right),\sin \theta = 2\sin \left( {\frac{\theta }{2}} \right)\cos \left( {\frac{\theta }{2}} \right)$
$Let\quad x = \tan \theta \Rightarrow \theta = {\tan ^{ - 1}}x$ ${\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x}$ $= {\tan ^{ - 1}}\frac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \theta }}$ $= {\tan ^{ - 1}}\frac{{\sqrt {{{\sec }^2}\theta } - 1}}{{\tan \theta }}$ $= {\tan ^{ - 1}}\frac{{\sec \theta - 1}}{{\tan \theta }}$ $= {\tan ^{ - 1}}\left( {\frac{{\frac{1}{{\cos \theta }} - 1}}{{\frac{{\sin \theta }}{{\cos \theta }}}}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)$ $= {\tan ^{ - 1}}\left( {\tan \frac{\theta }{2}} \right)$ $= \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x$

Question (6)

${\tan ^{ - 1}}\frac{1}{{\sqrt {{x^2} - 1} }},|x| > 1$

Solution

The trigonometric formulas to be used $\tan \left( {\frac{\pi }{2} - \theta } \right) = \cot \theta$
$let\;x = \sec \theta \Rightarrow \theta = {\sec ^{ - 1}}x$ ${\tan ^{ - 1}}\frac{1}{{\sqrt {{x^2} - 1} }}$ $= {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt {{{\sec }^2}\theta - 1} }}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{1}{{\tan \theta }}} \right)$ $= {\tan ^{ - 1}}\left( {\cot \theta } \right)$ $= {\tan ^{ - 1}}\left[ {\tan \left( {\frac{\pi }{2} - \theta } \right)} \right]$ $= \left( {\frac{\pi }{2} - \theta } \right)$ $= \left( {\frac{\pi }{2} - {{\sec }^{ - 1}}x} \right)$

Question (7)

${\tan ^{ - 1}}\left( {\sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} } \right),x < \pi$

Solution

The trigonometric formulas to be used $1 - \cos \theta = 2{\sin ^2}\left( {\frac{\theta }{2}} \right),1 + \cos \theta = 2{\cos ^2}\left( {\frac{\theta }{2}} \right)$
${\tan ^{ - 1}}\left( {\sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} } \right)$ $= {\tan ^{ - 1}}\left( {\sqrt {\frac{{2{{\sin }^2}\left( {\frac{x}{2}} \right)}}{{2{{\cos }^2}\left( {\frac{x}{2}} \right)}}} } \right)$ $= {\tan ^{ - 1}}\sqrt {{{\tan }^2}\left( {\frac{x}{2}} \right)}$ $= {\tan ^{ - 1}}\left( {\tan \frac{x}{2}} \right)$ $= \frac{x}{2}$

Question (8)

${\tan ^{ - 1}}\left( {\frac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right),0 < x < \pi$

Solution

The formula to be used ${\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}x - {\tan ^{ - 1}}y$
${\tan ^{ - 1}}\left( {\frac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right)$ Divide by cos x to numerator and denominator.
$= {\tan ^{ - 1}}\left( {\frac{{\frac{{\cos x}}{{\cos x}} - \frac{{\sin x}}{{\cos x}}}}{{\frac{{\cos x}}{{\cos x}} + \frac{{\sin x}}{{\cos x}}}}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{1 - \tan x}}{{1 + \tan x}}} \right)$ $= {\tan ^{ - 1}}1 - {\tan ^{ - 1}}\left( {\tan x} \right)$ $= \frac{\pi }{4} - x$

Question (9)

${\tan ^{ - 1}}\frac{x}{{\sqrt {{a^2} - {x^2}} }},|x| < a$

Solution

The formula to be used $1 - {\sin ^2}\theta = {\cos ^2}\theta$
$letx = a\sin \theta \Rightarrow \theta {\sin ^{ - 1}}\frac{x}{a}$ ${\tan ^{ - 1}}\frac{x}{{\sqrt {{a^2} - {x^2}} }}$ $= {\tan ^{ - 1}}\left( {\frac{{a\sin \theta }}{{\sqrt {{a^2} - {a^2}{{\sin }^2}\theta } }}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{a\sin \theta }}{{a\cos \theta }}} \right)$ $= {\tan ^{ - 1}}\left( {\tan \theta } \right)$ $= \theta = {\sin ^{ - 1}}\left( {\frac{x}{a}} \right)$

Question (10)

${\tan ^{ - 1}}\left( {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),a > 0,\frac{{ - a}}{{\sqrt 3 }} \le x \le \frac{a}{{\sqrt 3 }}$

Solution

The formula to be used $\tan (3\theta ) = \frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}$
$let\;x = a\tan \theta \Rightarrow \theta = {\tan ^{ - 1}}\left( {\frac{x}{a}} \right)$ $\frac{{ - a}}{{\sqrt 3 }} \le x \le \frac{a}{{\sqrt 3 }} \Rightarrow \frac{{ - a}}{{\sqrt 3 }} \le a\tan \theta \le \frac{a}{{\sqrt 3 }}$ $\frac{{ - 1}}{{\sqrt 3 }} \le \tan \theta \le \frac{1}{{\sqrt 3 }}$ $- \tan \frac{\pi }{6} \le \tan \theta \le \frac{\pi }{6}$ $- \frac{\pi }{6} \le \theta \le \frac{\pi }{6}$ $- \frac{\pi }{2} \le 3\theta \le \frac{\pi }{2},3\theta \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$ ${\tan ^{ - 1}}\left( {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{3{a^2}a\tan \theta - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3a{a^2}{{\tan }^2}\theta }}} \right)$ $= {\tan ^{ - 1}}\left( {\frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right)$ $= {\tan ^{ - 1}}\left( {\tan 3\theta } \right)$ $= 3\theta = 3{\tan ^{ - 1}}\left( {\frac{x}{a}} \right)$ Find the values of each of the following:

Question (11)

${\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\frac{1}{2}} \right)} \right]$

Solution

${\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\frac{1}{2}} \right)} \right]$ $= {\tan ^{ - 1}}\left[ {2\cos \left( {2\frac{\pi }{6}} \right)} \right]$ $= {\tan ^{ - 1}}\left[ {2\cos \frac{\pi }{3}} \right]$ $= {\tan ^{ - 1}}\left( {2 \times \frac{1}{2}} \right)$ $= {\tan ^{ - 1}}1 = \frac{\pi }{4}$

Question (12)

$\cot \left( {{{\tan }^{ - 1}}a + {{\cot }^{ - 1}}a} \right)$

Solution

${\tan ^{ - 1}}a + {\cot ^{ - 1}}a = \frac{\pi }{2}$
$\cot \left( {{{\tan }^{ - 1}}a + {{\cot }^{ - 1}}a} \right)$ $= \cot \frac{\pi }{2} = 0$

Question (13)

$\tan \frac{1}{2}\left[ {{{\sin }^{ - 1}}\frac{{2x}}{{1 + {x^2}}} + {{\cos }^{ - 1}}\frac{{1 - {y^2}}}{{1 + {y^2}}}} \right],|x| < 1,y > 0\;and\;xy < 1$

Solution

The formula to be used. ${\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \sin 2\theta ,\;\;\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos 2\theta }$ $\tan \left( {\alpha + \beta } \right) = \frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}$
$Let\quad x = \tan \alpha ,\quad y = \tan \beta$ $\tan \frac{1}{2}\left[ {{{\sin }^{ - 1}}\frac{{2x}}{{1 + {x^2}}} + {{\cos }^{ - 1}}\frac{{1 - {y^2}}}{{1 + {y^2}}}} \right]$ $= \tan \frac{1}{2}\left[ {{{\sin }^{ - 1}}\frac{{2\tan \alpha }}{{1 + {{\tan }^2}\alpha }} + {{\cos }^{ - 1}}\frac{{1 - {{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }}} \right]$ $= \tan \frac{1}{2}\left[ {{{\sin }^{ - 1}}\left( {\sin 2\alpha } \right) + {{\cos }^{ - 1}}\left( {\cos 2\beta } \right)} \right]$ $= \tan \frac{1}{2}\left[ {2\alpha + 2\beta } \right]$ $= \tan \left( {\alpha + \beta } \right)$ $= \frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}$ $= \frac{{x + y}}{{1 - xy}}$

Question (14)

$If\sin \left[ {{{\sin }^{ - 1}}\frac{1}{5} + {{\cos }^{ - 1}}x} \right] = 1,$ then find the value of x.

Solution

$\sin \left[ {{{\sin }^{ - 1}}\frac{1}{5} + {{\cos }^{ - 1}}x} \right] = 1$ ${\sin ^{ - 1}}\frac{1}{5} + {\cos ^{ - 1}}x = {\sin ^{ - 1}}1$ ${\sin ^{ - 1}}\frac{1}{5} + {\cos ^{ - 1}}x = \frac{\pi }{2}$ ${\cos ^{ - 1}}x = \frac{\pi }{2} - {\sin ^{ - 1}}\frac{1}{5}$ $x = \cos \left( {\frac{\pi }{2} - {{\sin }^{ - 1}}\frac{1}{5}} \right)$ $x = \sin \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)$ $x = \frac{1}{5}$

Question (15)

$If{\tan ^{ - 1}}\frac{{x - 1}}{{x - 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4},$ then find the value of x.

Solution

${\tan ^{ - 1}}\frac{{x - 1}}{{x - 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4}$ ${\tan ^{ - 1}}\left[ {\frac{{\frac{{x - 1}}{{x - 2}} + \frac{{x + 1}}{{x + 2}}}}{{1 - \left( {\frac{{x - 1}}{{x - 2}} \times \frac{{x + 1}}{{x + 2}}} \right)}}} \right] = \frac{\pi }{4}$ ${\tan ^{ - 1}}\left[ {\frac{{\left( {x - 1} \right)\left( {x + 2} \right) + \left( {x + 1} \right)\left( {x - 2} \right)}}{{{x^2} - 4 - \left( {{x^2} - 1} \right)}}} \right] = \frac{\pi }{4}$ $\frac{{{x^2} + x - 2 + {x^2} - x - 2}}{{{x^2} - 4 - {x^2} + 1}} = \tan \frac{\pi }{4}$ $\frac{{2{x^2} - 4}}{{ - 3}} = 1$ $2{x^2} - 4 = - 3$ $2{x^2} = - 3 + 4 = 1$ ${x^2} = \frac{1}{2} \Rightarrow x = \pm \frac{1}{{\sqrt 2 }}$ Find the values of each of the following:

Question (16)

${\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{3}} \right)$

Solution

${\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta ,when\theta \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$
${\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{3}} \right)$ $= {\sin ^{ - 1}}\left( {\sin \left( {\pi - \frac{\pi }{3}} \right)} \right)$ $= {\sin ^{ - 1}}{\rm{(}}\sin \frac{\pi }{3})$ $= \frac{\pi }{3}$

Question (17)

${\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right)$

Solution

${\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta ,when\theta \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$
${\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right)$ $= {\tan ^{ - 1}}\left( {\tan \left( {\pi - \frac{\pi }{4}} \right)} \right)$ $= {\tan ^{ - 1}}\left( { - \tan \frac{\pi }{4}} \right)$ $= - {\tan ^{ - 1}}\left( {\tan \frac{\pi }{4}} \right)$ $= - \frac{\pi }{4}$

Question (18)

$\tan \left( {{{\sin }^{ - 1}}\frac{3}{5} + {{\cot }^{ - 1}}\frac{3}{2}} \right)$

Solution

The formula to be used ${\sin ^{ - 1}}x = {\tan ^{ - 1}}\frac{x}{{\sqrt {1 - {x^2}} }},\;{\cot ^{ - 1}}x = {\tan ^{ - 1}}\frac{1}{x}$
$\tan \left( {{{\sin }^{ - 1}}\frac{3}{5} + {{\cot }^{ - 1}}\frac{3}{2}} \right)$ $= \tan \left( {{{\tan }^{ - 1}}\frac{3}{4} + {{\tan }^{ - 1}}\frac{2}{3}} \right)$ $= \tan \left( {{{\tan }^{ - 1}}\frac{{\frac{3}{4} + \frac{2}{3}}}{{1 - \left( {\frac{3}{4} \times \frac{2}{3}} \right)}}} \right)$ $= \tan \left( {{{\tan }^{ - 1}}\left( {\frac{{\frac{{9 + 8}}{{12}}}}{{\frac{{12 - 6}}{{12}}}}} \right)} \right)$ $= \tan \left( {{{\tan }^{ - 1}}\frac{{17}}{6}} \right) = \frac{{17}}{6}$

Question (19)

${\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right)$ (A) $\frac{{7\pi }}{6}$   (B)$\frac{{5\pi }}{6}$   (C) $\frac{\pi }{3}$   (D) $\frac{\pi }{6}$

Solution

${\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta ,\theta \in \left[ {0,\pi } \right]$
${\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right)$ $= {\cos ^{ - 1}}\left( {\cos \left( {\pi + \frac{\pi }{6}} \right)} \right)$ $= {\cos ^{ - 1}}\left( { - \cos \frac{\pi }{6}} \right)$ $= \pi - {\cos ^{ - 1}}\left( {\cos \frac{\pi }{6}} \right)$ $= \pi - \frac{\pi }{6} = \frac{{5\pi }}{6}$ So B is the correct answer.

Question (20)

$\sin \left[ {\frac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \frac{1}{2}} \right)} \right]$ (A)1/2   (B) 1/3   (C)1/4   (D) 1

Solution

$\sin \left[ {\frac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \frac{1}{2}} \right)} \right]$ $= \sin \left[ {\frac{\pi }{3} + {{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)} \right]$ $= \sin \left[ {\frac{\pi }{3} + \frac{\pi }{6}} \right]$ $= \sin \frac{\pi }{2} = 1$ So D is the correct option.

Question (21)

${\tan ^{ - 1}}\sqrt 3 - {\cot ^{ - 1}}\left( { - \sqrt 3 } \right)is\;equal\;to$ (A) $\pi$   (B) $- \frac{\pi }{2}$   (C) 0   (D)$2\sqrt 3$

Solution

${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$
${\tan ^{ - 1}}\sqrt 3 - {\cot ^{ - 1}}\left( { - \sqrt 3 } \right)$ $= {\tan ^{ - 1}}\sqrt 3 - \left( {\pi - {{\cot }^{ - 1}}\left( {\sqrt 3 } \right)} \right)$ $= {\tan ^{ - 1}}\sqrt 3 - \pi + {\cot ^{ - 1}}\left( {\sqrt 3 } \right)$ $= \frac{\pi }{2} - \pi = - \frac{\pi }{2}$ So B is the correct option.