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# 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:

AC^{2} = AB^{2} + BC^{2}

BC^{2} = AC^{2} - AB^{2}

BC^{2} = (13k)^{2} - (12k)^{2}

BC^{2} = 169 k^{2} - 144 k^{2}

BC^{2} = 25k^{2}

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

**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.

**☛ Related Questions:**

- If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
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- If 3 cot A = 4, check whether (1 - tan2 A) / (1 + tan2 A) = cos2 A - sin2 A or not.
- In the triangle ABC right-angled at B, if tan A = 1/√3 find the value of:(i) sin A cos C + cos A sin C(ii) cos A cos C - sin A sin C

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