Given sec θ = 13/12, calculate all other trigonometric ratios

Solution:

We will use the basic concept of trigonometric ratios to solve the problem.

Let ΔABC be a right-angled triangle, right angled at point B.

It is given that:

sec θ = hypotenuse / side adjacent to ∠θ = AC/AB = 13/12

Let AC = 13k and AB = 12k, where k is a positive integer.

Applying Pythagoras theorem in Δ ABC, we obtain:

AC² = AB² + BC²

BC² = AC² - AB²

BC² = (13k)² - (12k)²

BC² = 169 k² - 144 k²

BC² = 25k²

BC = 5k

sin θ = side opposite to ∠θ / hypotenuse = BC/AC = 5/13

cos θ = side adjacent to ∠θ / hypotenuse = AB/AC = 12/13

tan θ = side opposite to ∠θ / side adjacent to ∠θ = BC/AB = 5/12

cot θ = side adjacent to ∠θ / side opposite to ∠θ = AB/BC = 12/5

cosec θ = hypotenuse / side opposite to ∠θ = AC/BC = 13/5

☛ Check: NCERT Solutions Class 10 Maths Chapter 8

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Video Solution:

Given sec θ = 13/12, calculate all other trigonometric ratios.

Maths NCERT Solutions Class 10  Chapter 8 Exercise 8.1 Question 5

Summary:

If sec θ = 13/12, all the other trigonometric ratios are as follows: sin θ = 5/13, cos θ = 12/13, tan θ  = 5/12, cot θ  = 12/5 and cosec θ = 13/5 respectively.

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