$\tan \theta=\sqrt{3},$ તો $\theta=\ldots \ldots \ldots$
$30$
$45$
$60$
$90$
$\tan \theta=\sqrt{3} \cdot$ But, $\tan 60=\sqrt{3}\, \therefore \theta=60$
સાબિત કરો :
$(\sqrt{3}+ 1) \left(3-\cot 30^{\circ}\right)=\tan ^{3} 60^{\circ}-2 \sin 60^{\circ}$
$\Delta ABC$ માં , $m \angle B =90, BC =3$ અને $AC =5,$ તો $\tan A =\ldots \ldots \ldots \ldots$
$\sin ^{2} 35+\cos ^{2} \theta=1,$ તો $\theta=\ldots \ldots \ldots \ldots$
$\tan ^{2} \theta-\sec ^{2} \theta=\ldots \ldots \ldots$
$a \sin \theta=3$ અને $a \cos \theta=4,$ તો $a=\ldots \ldots \ldots . . .$ (કે જ્યાં, $a > 0)$
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