If sin θ = $\dfrac{1}{\sqrt{2}}$, find the values of other trigonometrical ratios for θ.

sin θ = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{2}}$

Let perpendicular = x and hypotenuse = $\sqrt{2}x$

By using Pythagoras theorem, we get :

Hypotenuse² = Base² + Perpendicular²

Base² = Hypotenuse² - Perpendicular²

Base² = $(\sqrt{2}x)^2$ - x²

Base² = 2x² - x²

Base² = x²

Base = x

Now, calculating the remaining trigonometric ratios :

cos θ = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{x}{\sqrt{2}x} = \dfrac{1}{\sqrt{2}}$

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

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

sec θ = $\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{\sqrt{2}x}{x} = \sqrt{2}$

cosec θ = $\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{\sqrt{2}x}{x} = \sqrt{2}$