Delta XYZ is right-angled at Y. If cos X = fr | vedclass

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(B) In $\Delta XYZ$,$\angle Y = 90^\circ$.Given $\cos X = \frac{3}{5}$.By definition,$\cos X = \frac{	ext{Adjacent side to } X}{	ext{Hypotenuse}} =

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$\frac{5}{3}$

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(B) In $\Delta XYZ$,$\angle Y = 90^\circ$.Given $\cos X = \frac{3}{5}$.By definition,$\cos X = \frac{	ext{Adjacent side to } X}{	ext{Hypotenuse}} = \frac{XY}{XZ} = \frac{3}{5}$.Now,for $\angle Z$,the side $XY$ is the opposite side and $XZ$ is the hypotenuse.Therefore,$\sin Z = \frac{	ext{Opposite side to } Z}{	ext{Hypotenuse}} = \frac{XY}{XZ} = \frac{3}{5}$.We know that $\operatorname{cosec} Z = \frac{1}{\sin Z}$.Thus,$\operatorname{cosec} Z = \frac{1}{3/5} = \frac{5}{3}$.

$\Delta XYZ$ is right-angled at $Y$. If $\cos X = \frac{3}{5}$,then what is the value of $\operatorname{cosec} Z$?

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