Inverse trigonometric ratios: cosecant, secant and cotangent

In this section, we are going to define the inverse trigonometric ratios, this is, the inverse ratios of the sine, the cosine and the tangent. Given a triangle rectangle, we define the cosecant, the secant and the cotangent of an angle $x$ as the inverse ratios of the sine, the cosine and the tangent, respectively.

• $\csc(x)$: the cosecant is the inverse of the sine or, also, its multiplicative inverse: $$\csc(x)=\dfrac{1}{\sin(x)}=\dfrac{c}{a}$$

• $\sec (x)$: the secant is the inverse of the cosine or, also, its multiplicative inverse: $$\sec(x)=\dfrac{1}{\cos(x)}=\dfrac{c}{b}$$

• $\cot(x)$: the cotangent is the inverse of the tangent or, also, its multiplicative inverse: $$\cot(x)=\dfrac{1}{\tan(x)}=\dfrac{b}{a}$$

Given the triangle of sides $a = 3$, $b = 4$ and $c = 5$, we are going to compute the trigonometric ratios associated with such a triangle.

Then: $$\sin(x)= \dfrac{3}{5} \qquad \cos(x)=\dfrac{4}{5} \qquad \tan(x)=\dfrac{3}{4}$$

The associated inverse trigonometric ratios are: $$\csc(x)= \dfrac{5}{3} \qquad \sec(x)=\dfrac{5}{4} \qquad \cot(x)=\dfrac{4}{3}$$

Given the triangle of sides $a = 5$, $b = 12$ and $c = 13$, compute its trigonometric ratios.

$$\sin(x)= \dfrac{5}{13}= \qquad \cos(x)=\dfrac{12}{13} \qquad \tan(x)=\dfrac{5}{12}$$

$$\csc(x)= \dfrac{13}{5} \qquad \sec(x)=\dfrac{13}{12} \qquad \cot(x)=\dfrac{12}{5}$$