MCQ
${\sin ^{ - 1}}\frac{{\sqrt x }}{{\sqrt {x + a} }} =\ . .... ..$
  • A
    ${\cos ^{ - 1}}\sqrt {\frac{x}{a}} $
  • B
    ${\rm{cose}}{{\rm{c}}^{ - 1}}\sqrt {\frac{x}{a}} $
  • ${\tan ^{ - 1}}\sqrt {\frac{x}{a}} $
  • D
    એકપણ નહીં.

Answer

Correct option: C.
${\tan ^{ - 1}}\sqrt {\frac{x}{a}} $
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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