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2. inverse trigonometric function — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths2. inverse trigonometric function1 Mark
MCQ
If $x$  takes non-positive permissible value, then ${\sin ^{ - 1}}x =$
  • A
    ${\cos ^{ - 1}}\sqrt {1 - {x^2}} $
  • ✓
    $ - {\cos ^{ - 1}}\sqrt {1 - {x^2}} $
  • C
    ${\cos ^{ - 1}}\sqrt {{x^2} - 1} $
  • D
    $\pi - {\cos ^{ - 1}}\sqrt {1 - {x^2}} $
✓

Answer

Correct option: B.
$ - {\cos ^{ - 1}}\sqrt {1 - {x^2}} $
b
(b) Let ${\sin ^{ - 1}}x = y.$ Then $x = \sin y$

Since $ - 1 \le x \le 0,$   therefore $\frac{{ - \pi }}{2} \le {\sin ^{ - 1}}x \le 0$ 

and so $\frac{{ - \pi }}{2} \le y \le 0$

We have $\cos y = \sqrt {1 - {{\sin }^2}y} $

$ \Rightarrow \,\,\cos y = \sqrt {1 - {x^2}} $, for $0 \le y \le \pi $   …..$(i)$

Now $ - \frac{\pi }{2} \le y \le 0\,\, \Rightarrow \,\,\frac{\pi }{2} \ge - y \ge 0$

$ \Rightarrow \,\,\cos \,\left( { - y} \right) = \sqrt {1 - {x^2}} $     {from $(i)$}

$ \Rightarrow \,\, - y = {\cos ^{ - 1}}\sqrt {1 - {x^2}} \,\, $

$\Rightarrow \,\,y = - {\cos ^{ - 1}}\sqrt {1 - {x^2}} $.

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