12th NCERT Inverse Trigonometric Functions Exercise 2.1 Questions 14
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Find the principle values of the following:

Question (1)

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

Solution

\[{\sin ^{ - 1}}( - x) = - {\sin ^{ - 1}}x\]
\[{\sin ^{ - 1}}\left( { - \frac{1}{2}} \right)\] \[ = - {\sin ^{ - 1}}\left( {\frac{1}{2}} \right)\] \[ = - \frac{\pi }{6}\]

Question (2)

\[{\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)\]

Solution

\[{\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)\] \[ = {\rm{ }}\frac{\pi }{6}\]

Question (3)

\[\cos e{c^{ - 1}}(2)\]

Solution

\[\cos e{c^{ - 1}}(x) = {\sin ^{ - 1}}\frac{1}{x}\]
\[\cos e{c^{ - 1}}(2)\] \[ = {\rm{ }}{\sin ^{ - 1}}\left( {\frac{1}{2}} \right)\] \[ = {\rm{ }}\frac{\pi }{6}\]

Question (4)

\[{\tan ^{ - 1}}\left( { - \sqrt 3 } \right)\]

Solution

\[{\tan ^{ - 1}}( - x) = - {\tan ^{ - 1}}x\]
\[{\tan ^{ - 1}}\left( { - \sqrt 3 } \right)\] \[ - {\rm{ }}{\tan ^{ - 1}}\left( {\sqrt 3 } \right)\] \[ - {\rm{ }}\frac{\pi }{3}\]

Question (5)

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

Solution

\[{\cos ^{ - 1}}( - x) = \pi - {\cos ^{ - 1}}x\]
\[{\cos ^{ - 1}}\left( { - \frac{1}{2}} \right)\] \[ = \pi - {\rm{ }}{\cos ^{ - 1}}\left( {\frac{1}{2}} \right)\] \[ = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3}\]

Question (6)

\[{\tan ^{ - 1}}\left( { - 1} \right)\]

Solution

\[{\tan ^{ - 1}}( - x) = - {\tan ^{ - 1}}x\]
\[{\tan ^{ - 1}}\left( { - 1} \right)\] \[ = - {\rm{ }}{\tan ^{ - 1}}\left( 1 \right)\] \[ = - {\rm{ }}\frac{\pi }{4}\]

Question (7)

\[{\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)\]

Solution

\[{\sec ^{ - 1}}(x) = {\cos ^{ - 1}}\frac{1}{x}\]
\[{\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)\] \[ = {\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)\] \[ = \frac{\pi }{6}\]

Question (8)

\[{\cot ^{ - 1}}\left( {\sqrt 3 } \right)\]

Solution

\[{\cot ^{ - 1}}\left( {\sqrt 3 } \right)\] \[ = \frac{\pi }{6}\]

Question (9)

\[{\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right)\]

Solution

\[{\cos ^{ - 1}}( - x) = \pi - {\cos ^{ - 1}}x\]
\[{\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right)\] \[ = \pi - {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right)\] \[ = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}\]

Question (10)

\[\;\cos e{c^{ - 1}}\left( { - \sqrt 2 } \right)\]

Solution

\[co{\sec ^{ - 1}}( - x) = - \cos e{c^{ - 1}}x\]
\[\;\cos e{c^{ - 1}}\left( { - \sqrt 2 } \right)\] \[ = - \;\cos e{c^{ - 1}}\left( {\sqrt 2 } \right)\] \[ = - \frac{\pi }{4}\] Find the values of the followings

Question (11)

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

Solution

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

Question (12)

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

Solution

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

Question (13)

\[If\;\;{\sin ^{ - 1}}x = y,then\;\] (a) \[0 \le y \le \pi \]   (b) \[ - \frac{\pi }{2} \le y \le \frac{\pi }{2}\]   (c) \[0 < y < \pi \]   (d) \[ - \frac{\pi }{2} < y < \frac{\pi }{2}\]

Solution

\[{\sin ^{ - 1}}x = y\] y is the range of sin -1. And its range is \[ - \frac{\pi }{2} \le y \le \frac{\pi }{2}\] So B is the correct option.

Question (14)

\[{\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {\sec ^{ - 1}}\left( { - 2} \right)\;is\;equal\;to\] (a) \[\pi \]   (b) \[ - \frac{\pi }{3}\]   (c) \[\frac{\pi }{3}\]   (d) \[\frac{{2\pi }}{3}\]

Solution

\[{\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {\sec ^{ - 1}}\left( { - 2} \right)\] \[\; = \frac{\pi }{3} - \left( {\pi - {{\sec }^{ - 1}}\left( 2 \right)} \right)\] \[ = \frac{\pi }{3} - \left( {\pi - \frac{\pi }{3}} \right)\] \[ = \frac{\pi }{3} - \frac{{2\pi }}{3} = - \frac{\pi }{3}\] So B is the correct option.
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⇒ Exercise 2.2