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}} $
- DNone of these
${\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)$.
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$R =\left\{( x , y ): 5 x ^{2} \leq y \leq 2 x ^{2}+9\right\}$ is ........ $square\, units$