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

Gujarat BoardEnglish MediumSTD 12 ScienceMaths2. inverse trigonometric function1 Mark
MCQ
${\sin ^{ - 1}}\frac{{\sqrt x }}{{\sqrt {x + a} }}$ is equal to
  • A
    ${\cos ^{ - 1}}\sqrt {\frac{x}{a}} $
  • B
    ${\rm{cose}}{{\rm{c}}^{ - 1}}\sqrt {\frac{x}{a}} $
  • ✓
    ${\tan ^{ - 1}}\sqrt {\frac{x}{a}} $
  • D
    None of these
✓

Answer

Correct option: C.
${\tan ^{ - 1}}\sqrt {\frac{x}{a}} $
c
(c) Putting $x = a\,{\tan ^2}\theta $

${\sin ^{ - 1}}\frac{{\sqrt x }}{{\sqrt {x + a} }} = {\sin ^{ - 1}}\frac{{\sqrt a \sqrt {{{\tan }^2}\theta } }}{{\sqrt {a\,{{\tan }^2}\theta + a} }} $

$= {\sin ^{ - 1}}\frac{{\sqrt a \,\tan \theta }}{{\sqrt a \,\sec \theta }}$

$ = {\sin ^{ - 1}}\sin \theta = \theta = {\tan ^{ - 1}}\left( {\sqrt {\frac{x}{a}} } \right)$.

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