# Trigonometric Ratios

Let there be a right triangle ABC, right angled at B. Here ∠A (∠CAB) is an acute angle, AC is hypotenuse, side BC is opposite to ∠A and side AB is adjacent to ∠A.

If you consider acute ∠C, then side AB is side opposite to ∠C and side BC is adjacent to ∠C.

The trigonometric ratios of ∠A in right angled ΔABC are defined as:

$$ \sin A = \frac{BC}{AC} $$

$$ \cos A = \frac{AB}{AC} $$

$$ \tan A = \frac{BC}{AB} $$

$$ \csc A = \frac{AC}{BC} $$

$$ \sec A = \frac{AC}{AB} $$

$$ \cot A = \frac{AB}{BC} $$

When two sides of a right-triangle are given, its third side can be found out by using the Pythagoras theorem. Then, you can find the trigonometric ratios of the given angle.

Sometimes you know one trigonometric ratio and you have to find the values of other t-ratios. This can be easily done by using the definition of t-ratios and the Pythagoras Theorem.

### Relationships Between T-Ratios

$$ \tan \theta = \frac{\sin \theta}{\cos \theta} $$

**cosec θ is the reciprocal of sin θ**

$$ \csc \theta = \frac{1}{\sin \theta} $$

**sec θ is reciprocal of cos θ**

$$ \sec \theta = \frac{1}{\cos \theta} $$

**cot θ is reciprocal of tan θ**

$$ \cot \theta = \frac{1}{\tan \theta} $$

$$ \cot \theta = \frac{\cos \theta}{\sin \theta} $$

Thus, cosec θ, sec θ and cot θ are reciprocal of sin θ, cos θ and tan θ respectively.