12th NCERT Inverse Trigonometric Functions Exercise 2.2 Questions 21
Do or do not
There is no try

Prove the following :

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.
Exercise 2.1 ⇐
⇒ Miscallneous