From the figure (1) given below, find the values of:

(i) sin θ

(ii) cos θ

(iii) tan θ

(iv) cot θ

(v) sec θ

(vi) cosec θ

In right-angled triangle OMP,

By Pythagoras theorem, we get

⇒ OP² = OM² + MP²

⇒ MP² = OP² - OM²

⇒ MP² = (15)² - (12)²

⇒ MP² = 225 - 144

⇒ MP² = 81

⇒ MP = $\sqrt{81}$ = 9.

(i) sin θ = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{MP}{OP} = \dfrac{9}{15} = \dfrac{3}{5}$.

Hence, sin θ = $\dfrac{3}{5}$.

(ii) cos θ = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

= $\dfrac{OM}{OP} = \dfrac{12}{15} = \dfrac{4}{5}$.

Hence, cos θ = $\dfrac{4}{5}$.

(iii) tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}}$

= $\dfrac{MP}{OM} = \dfrac{9}{12} = \dfrac{3}{4}$.

Hence, tan θ = $\dfrac{3}{4}$.

(iv) cot θ = $\dfrac{\text{Base}}{\text{Perpendicular}}$

= $\dfrac{OM}{MP} = \dfrac{12}{9} = \dfrac{4}{3}$.

Hence, cot θ = $\dfrac{4}{3}$.

(v) sec θ = $\dfrac{\text{Hypotenuse}}{\text{Base}}$

= $\dfrac{OP}{OM} = \dfrac{15}{12} = \dfrac{5}{4}$.

Hence, sec θ = $\dfrac{5}{4}$.

(vi) cosec θ = $\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}$

= $\dfrac{OP}{MP} = \dfrac{15}{9} = \dfrac{5}{3}$.

Hence, cosec θ = $\dfrac{5}{3}$.