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.