$\sec (90^\circ - \theta) = \dots$

  • A
    $\cos \theta$
  • B
    $\operatorname{cosec} \theta$
  • C
    $\tan \theta$
  • D
    $\cot \theta$

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Similar Questions

$\cos (40^{\circ}-\theta)-\sin (50^{\circ}+\theta) = \ldots \ldots \ldots \ldots$

Prove that,$\tan \theta + \tan (90^{\circ} - \theta) = \sec \theta \sec (90^{\circ} - \theta)$

$\cos \theta = \frac{b}{\sqrt{a^2 + b^2}}$; where,$0 < \theta < 90^\circ$; then $\sin \theta = \dots$

If $a \sin \theta = 3$ and $a \cos \theta = 4$,then $a = \dots$ (where $a > 0$).

Prove that $(\sin^{4} \theta - \cos^{4} \theta + 1) \operatorname{cosec}^{2} \theta = 2$.

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