Given sec θ = \(\dfrac{13}{12}\), calculate all other trigonometric ratios.
sin θ = \(\dfrac{5}{13}\), cos θ = \(\dfrac{12}{13}\), tan θ = \(\dfrac{5}{12}\), cot θ = \(\dfrac{12}{5}\), cosec θ = \(\dfrac{13}{5}\)
Given: \(\sec\theta = \dfrac{13}{12}\)
To find: \(\sin\theta,\ \cos\theta,\ \tan\theta,\ \cot\theta,\ \text{cosec }\theta\).
Step 1: Use the definition of secant
We know:
\(\sec\theta = \dfrac{1}{\cos\theta} = \dfrac{\text{hypotenuse}}{\text{adjacent}}\)
So, if \(\sec\theta = \dfrac{13}{12}\), we can take:
Hypotenuse \(= 13\) and Adjacent \(= 12\) (with respect to angle \(\theta\)).
Student Note: Whenever a ratio is given as a fraction, you can treat numerator and denominator as sides of a right triangle (same scale). This makes finding other ratios easy.
Step 2: Find the opposite side using Pythagoras theorem
In a right triangle:
\((\text{hypotenuse})^2 = (\text{adjacent})^2 + (\text{opposite})^2\)
So,
\(13^2 = 12^2 + (\text{opposite})^2\)
\(169 = 144 + (\text{opposite})^2\)
\((\text{opposite})^2 = 169 - 144 = 25\)
\(\text{opposite} = \sqrt{25} = 5\).
Quick Triangle Summary: Adjacent = \(12\), Opposite = \(5\), Hypotenuse = \(13\).
Step 3: Write all required trigonometric ratios
(a) Cosine:
\(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{12}{13}\)
(b) Sine:
\(\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{5}{13}\)
(c) Tangent:
\(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{5}{12}\)
(d) Cotangent:
\(\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{12}{5}\)
(e) Cosecant:
\(\text{cosec }\theta = \dfrac{1}{\sin\theta} = \dfrac{13}{5}\)
Student Note (Check): Since \(\sec\theta = \dfrac{1}{\cos\theta}\), and we got \(\cos\theta = \dfrac{12}{13}\), its reciprocal is \(\dfrac{13}{12}\) which matches the given value. So everything is consistent.