If cosec θ = $\sqrt{10}$, find the values of other trigonometrical ratios for θ.

cosec θ = $\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{\sqrt{10}}{1}$.

Let hypotenuse = $\sqrt{10}x$ and perpendicular = x.

By pythagoras theorem, we get :

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

Base² = $(\sqrt{10}x)^2$ - x²

Base² = 10x² - x²

Base² = 9x²

Base = $(\sqrt{9x^2})$

Base = 3x

sin θ = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{x}{\sqrt{10}x} = \dfrac{1}{\sqrt{10}}$

cos θ = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{3x}{\sqrt{10}x} = \dfrac{3}{\sqrt{10}}$

tan θ = $\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{x}{3x} = \dfrac{1}{3}$

cot θ = $\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{3x}{x} = 3$

sec θ = $\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{\sqrt{10}x}{3x} = \dfrac{\sqrt{10}}{3}$