In Delta ABC,m angle C = 90^{circ} and cos B | vedclass

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(D) Given that in $\Delta ABC$,$m \angle C = 90^{\circ}$ and $\cos B = \frac{1}{2}$.We know that $\cos B = \frac{	ext{adjacent side to } \angle B}{

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$2$

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(D) Given that in $\Delta ABC$,$m \angle C = 90^{\circ}$ and $\cos B = \frac{1}{2}$.We know that $\cos B = \frac{	ext{adjacent side to } \angle B}{	ext{hypotenuse}} = \frac{BC}{AB}$.Therefore,$\frac{BC}{AB} = \frac{1}{2}$. Let $BC = k$ and $AB = 2k$,where $k > 0$.Using the Pythagorean theorem,$AC^2 + BC^2 = AB^2$.$AC^2 + k^2 = (2k)^2 = 4k^2$.$AC^2 = 3k^2 \implies AC = k\sqrt{3}$.Now,$\operatorname{cosec} A = \frac{	ext{hypotenuse}}{	ext{side opposite to } \angle A} = \frac{AB}{BC}$.$\operatorname{cosec} A = \frac{2k}{k} = 2$.

In $\Delta ABC$,$m \angle C = 90^{\circ}$ and $\cos B = \frac{1}{2}$,then $\operatorname{cosec} A = \ldots$

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