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.
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$$\csc(x)$$: the cosecant is the inverse of the sine or, also, its multiplicative inverse: $$$ \csc(x)=\dfrac{1}{\sin(x)}=\dfrac{c}{a}$$$
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$$\sec (x)$$: 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.
$$$ \sin(x)= \dfrac{5}{13}= \qquad \cos(x)=\dfrac{12}{13} \qquad \tan(x)=\dfrac{5}{12}$$$
$$$ \csc(x)= \dfrac{13}{5} \qquad \sec(x)=\dfrac{13}{12} \qquad \cot(x)=\dfrac{12}{5}$$$