Question

How do you find the remaining trigonometric functions of $\theta $ given $\cot \theta = \dfrac{m}{n}$ where $m\;{\text{and}}\;n$ are both positive?

Answer 1

Hint: To find the remaining trigonometric functions of $\theta $ given $\cot \theta = \dfrac{m}{n}$, use trigonometric relations, definitions and also trigonometric identities to find other trigonometric functions of $\theta $

Complete step by step solution:
In order to find the remaining trigonometric functions of $\theta $ given $\cot \theta = \dfrac{m}{n}$ where $m\;{\text{and}}\;n$ are both positive, we will find them with help of trigonometric relations, definitions and identities as follows:
There are six trigonometric functions which are Sine, Cosine, Tangent, Cosecant, Secant and Cotangent (one which is given) and they are represented as $\sin ,\;\cos ,\,\tan ,\;\csc ,\;\sec \;{\text{and}}\;\cot $ respectively. Among these trigonometric functions, Sine and Cosecant, Cosine and Secant, and Tangent and Cotangent are pairs which are multiplicative inverses of each other.
Now, from relations between trigonometric functions, we know that
$\tan x = \dfrac{1}{{\cot x}}$
So we can also write it as
$\tan \theta = \dfrac{1}{{\cot \theta }} = \dfrac{1}{{\dfrac{m}{n}}} = \dfrac{n}{m}$
Now, for finding values of the functions further, we will use the following trigonometric identity
${\sec ^2}x - {\tan ^2}x = 1$
We can further write it as
\[
  {\sec ^2}x = {\tan ^2}x + 1 \\
  {\text{or}} \\
  {\sec ^2}\theta = {\tan ^2}\theta + 1 \\
   \Rightarrow {\sec ^2}\theta = {\left( {\dfrac{n}{m}} \right)^2} + 1 \\
   \Rightarrow {\sec ^2}\theta = \dfrac{{{n^2}}}{{{m^2}}} + 1 \\
   \Rightarrow {\sec ^2}\theta = \dfrac{{{n^2} + {m^2}}}{{{m^2}}} \\
 \]
Taking square root both sides, we will get
\[
   \Rightarrow \sec \theta = \sqrt {\dfrac{{{n^2} + {m^2}}}{{{m^2}}}} \\
   \Rightarrow \sec \theta = \dfrac{1}{m}\sqrt {{n^2} + {m^2}} \\
 \]
So we get the value of secant, now again from the trigonometric relations, we know that
$\cos x = \dfrac{1}{{\sec x}}$
So we can also write it as
$\cos \theta = \dfrac{1}{{\sec \theta }} = \dfrac{1}{{\dfrac{1}{m}\sqrt {{n^2} + {m^2}} }} = \dfrac{m}{{\sqrt {{n^2} + {m^2}} }}$
Now, from the trigonometric definitions, we know that the tangent is the ratio of sine and cosine, that is
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Solving it for sine, we will get
\[
   \Rightarrow \sin \theta = \tan \theta \cos \theta \\
   \Rightarrow \sin \theta = \dfrac{n}{m} \times \dfrac{m}{{\sqrt {{n^2} + {m^2}} }} \\
   \Rightarrow \sin \theta = \dfrac{n}{{\sqrt {{n^2} + {m^2}} }} \\
 \]
Again from the trigonometric relations, we know that
$\csc x = \dfrac{1}{{\sin x}}$
So we can also write it as
\[\csc \theta = \dfrac{1}{{\sin \theta }} = \dfrac{1}{{\dfrac{n}{{\sqrt {{n^2} + {m^2}} }}}} = \dfrac{1}{n}\sqrt {{n^2} + {m^2}} \]
So we get the value of all trigonometric functions of $\theta $


Note: However the above pairs of the trigonometric functions are multiplicative inverse, even though there are some restrictions or you may say some arguments where these are not defined because their parent function then equals zero and give form $\dfrac{1}{0}$ which is not defined (e.g. $\sin \;{\text{and}}\;\csc \;{\text{at}}\;\theta = 0$)