Question

If $\text{cosec}A=$ $\sqrt{10}$, find other five trigonometric ratios

Answer 1

We have,

$\text{cosec}A=$ $\frac{Hypotenuse}{Perpendicular}=\frac{\sqrt{10}}{1}$

So, we draw a triangle ABC, right angled at B such that

Perpendicular $=BC=1$ unit and, Hypotenuse $=AC=$ $\sqrt{10}$ units.

By Pythagoras theorem, we have

$A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow (\sqrt{10}{)}^2=A{B}^2+1^2$

$\Rightarrow A{B}^2=10-1=9$

$\Rightarrow AB=\sqrt{9}=3$

When we consider the trigonometric ratios of $\angle$ A , we have

Base $=AB=3$, Perpendicular $=BC=1$, and Hypotenuse $=AC=$ $\sqrt{10}$

$\therefore sinA=\frac{Perpendicular}{Hypotenuse}=\frac{1}{\sqrt{10}}$

$cosA=\frac{Base}{Hypotenuse}=\frac{3}{\sqrt{10}}$

$tanA=\frac{Perpendicular}{Base}=\frac{1}{3}$

$secA=\frac{Hypotenuse}{Base}=\frac{\sqrt{10}}{3}$

and, $cotA=\frac{Base}{Perpendicular}=\frac{3}{1}=3$