Given that $\sin \theta = \frac{a}{b},$ then $\cos \theta$ is equal to

  • A
    $\frac{b}{\sqrt{b^{2}-a^{2}}}$
  • B
    $\frac{b}{a}$
  • C
    $\frac{a}{\sqrt{b^{2}-a^{2}}}$
  • D
    $\frac{\sqrt{b^{2}-a^{2}}}{b}$

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Similar Questions

If $\tan ^{2} \theta = \sin ^{2} \theta + \cos ^{2} \theta$,then $\theta = \ldots$ (in $^{\circ}$)

Given that $\alpha + \beta = 90^{\circ}$,show that $\sqrt{\cos \alpha \operatorname{cosec} \beta - \cos \alpha \sin \beta} = \sin \alpha$.

If $3 \theta$ is the measure of an acute angle and $\sin 3 \theta = \cos (\theta - 26^{\circ})$,then the value of $\theta$ is $\ldots \ldots \ldots \ldots$ (in $^{\circ}$)

Which of the following is true for some $\theta$ (where,$0 < \theta < 90^{\circ}$)?

$\frac{\sin 60^{\circ} + \cos 30^{\circ}}{1 + \sin 30^{\circ} + \cos 60^{\circ}} = \dots$

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