Class 11MathematicsTrigonometrical Ratios, Functions and IdentitiesFundamental trigonometrical ratios and functions, Trigonometrical ratio of allied angles
EasyIn a right-angled triangle $ABC$ where $\angle C = 90^{\circ}$,$AC = BC$. If $D$ is the midpoint of $AC$,then the cotangent of $\angle DBC$ is equal to:
- A$2$
- B$3$
- C$1/2$
- D$1/3$
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Similar Questions
The value of $\cos 1200^{\circ} + \tan 1485^{\circ}$ is
$\frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x} = ?$
Prove that:
$2 \sin ^{2} \frac{\pi}{6}+\csc ^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}$
$2 \sin ^{2} \frac{\pi}{6}+\csc ^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}$
Medium
View SolutionWhich of the following have the same value?
$(a)$ $\sin 120^{\circ}$
$(b)$ $\cos 930^{\circ}$
$(c)$ $\tan 840^{\circ}$
$(d)$ $\cot (-1110^{\circ})$
$(a)$ $\sin 120^{\circ}$
$(b)$ $\cos 930^{\circ}$
$(c)$ $\tan 840^{\circ}$
$(d)$ $\cot (-1110^{\circ})$
$\cot (45^\circ + \theta ) \cot (45^\circ - \theta ) = $
Easy
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