If $\sec \theta =\dfrac{13}{12}$ , then calculate all other trigonometric ratios.

Hint: Secant, cosecant and cotangent are the three additional functions which are derived from the primary functions of sine, cosine, and tangent. The reciprocal of sine, cosine, and tangent are cosecant, secant and cotangent respectively.

Complete step by step solution:
It is given that $\sec \theta =\dfrac{13}{12}$
We know that, $\sec \theta $ is a reciprocal of $\cos \theta $
$\cos \theta =\dfrac{1}{\sec \theta }=\dfrac{1}{\left( \dfrac{13}{12} \right)}=\dfrac{12}{13}$
We have $\cos \theta =\dfrac{12}{13}$
The cosine function of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse.
$\dfrac{\text{side adjacent to angle }\theta }{\text{Hypotenuse}}=\dfrac{12}{13}$
Let us assume that AB = 12x and AC = 13x
By using Pythagoras theorem to get side BC.
\[{{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Height} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}\]
${{\left( AC \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$
Now put all the values in the above equation, we get
${{\left( 13x \right)}^{2}}={{\left( 12x \right)}^{2}}+{{\left( BC \right)}^{2}}$
Rearranging the terms, we get
${{\left( BC \right)}^{2}}={{\left( 13x \right)}^{2}}-{{\left( 12x \right)}^{2}}=169{{x}^{2}}-144{{x}^{2}}=25{{x}^{2}}$
Taking the square root on both sides, we get
$BC=5x$

The sine function of an angle is the ratio between the opposite side length to that of the hypotenuse.
$\sin \theta =\dfrac{\text{side opposite to angle }\theta }{\text{Hypotenuse}}=\dfrac{BC}{AC}=\dfrac{5x}{13x}=\dfrac{5}{13}$
The tangent function is the ratio of the length of the opposite side to that of the adjacent side.
$\tan \theta =\dfrac{\text{side opposite to angle }\theta }{\text{side adjacent to angle }\theta }=\dfrac{BC}{AB}=\dfrac{5x}{12x}=\dfrac{5}{12}$
The $\cos ec\theta $ is the reciprocal of the $\sin \theta $ , we get
\[\cos ec\theta =\dfrac{1}{\sin \theta }=\dfrac{1}{\left( \dfrac{5}{13} \right)}=\dfrac{13}{5}\]
The $\cot \theta $ is the reciprocal of the $\tan \theta $ , we get
\[\cot \theta =\dfrac{1}{\tan \theta }=\dfrac{1}{\left( \dfrac{5}{12} \right)}=\dfrac{12}{5}\]
Hence \[\sin \theta =\dfrac{5}{13},\cos \theta =\dfrac{12}{13},\tan \theta =\dfrac{5}{12},\cos ec\theta =\dfrac{13}{5},\sec \theta =\dfrac{13}{12},\cot \theta =\dfrac{12}{5}\]

Note: Inverse trigonometric functions are used to obtain an angle from any of the angle’s trigonometric ratios. Basically, inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions are represented as arcsin, arccosine, arctangent, arccotangent, arcsecant, and arccosecant.