(A) Let $AC = BC = a$. Since $D$ is the midpoint of $AC$,$CD = a/2$.In the right-angled triangle $\triangle BCD$,$\angle C = 90^{\circ}$.Let $\angle D

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(A) Let $AC = BC = a$. Since $D$ is the midpoint of $AC$,$CD = a/2$.In the right-angled triangle $\triangle BCD$,$\angle C = 90^{\circ}$.Let $\angle DBC = \alpha$.Then,$\tan \alpha = \frac{CD}{BC} = \frac{a/2}{a} = \frac{1}{2}$.Therefore,$\cot \alpha = \frac{1}{\tan \alpha} = 2$.

Class 11 Mathematics Trigonometric Functions Fundamental trigonometrical ratios and functions, Trigonometrical ratio of allied angles

Easy

In 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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Trigonometric Functions

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Fundamental trigonometrical ratios and functions, Trigonometrical ratio of allied angles

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In 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:

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