Given sec θ = 13/12, calculate all other trigonometric ratios
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
Given sec θ = 13/12, calculate all other trigonometric ratios.
Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.1 Question 5
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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