Question

How do you prove $\dfrac{1-\tan \theta }{1+\tan \theta }=\dfrac{\cot \theta -1}{\cot \theta +1}$?

Answer 1

We first take the left sides of the equation and multiply $\cot \theta$ with both numerator and denominator. We know that $\cot \theta$ and $\tan \theta$ are trigonometric inverse to each other which gives $\tan \theta =\dfrac{1}{\cot \theta }$. The simplified form is $\tan \theta \cot \theta =1$. We replace the values and get the right side of the equation.

Complete step by step solution:
We take the left side equation and multiply $\cot \theta$ with both numerator and denominator.
We get $\dfrac{1-\tan \theta }{1+\tan \theta }\times \dfrac{\cot \theta }{\cot \theta }$.
The simplified form is $\dfrac{1-\tan \theta }{1+\tan \theta }\times \dfrac{\cot \theta }{\cot \theta }=\dfrac{\cot \theta -\tan \theta \cot \theta }{\cot \theta +\tan \theta \cot \theta }$.
As $\cot \theta$ and $\tan \theta$ are trigonometric inverse to each other, we can say that $\tan \theta \cot \theta =1$.
We replace the value in the expression to get
$\dfrac{1-\tan \theta }{1+\tan \theta }=\dfrac{\cot \theta -\tan \theta \cot \theta }{\cot \theta +\tan \theta \cot \theta }=\dfrac{\cot \theta -1}{\cot \theta +1}$.
Thus proved $\dfrac{1-\tan \theta }{1+\tan \theta }=\dfrac{\cot \theta -1}{\cot \theta +1}$.

Note:
We can show that $\cot \theta$ and $\tan \theta$ are trigonometric inverse to each other. With respect to the angle $\theta$, the $\cot \theta$ represents the ratio of the length of the base and height as $\cot \theta =\dfrac{base}{height}$ whereas $\tan \theta$ represents the ratio of the length of the height and base as $\tan \theta =\dfrac{height}{base}$. The multiplication of these two gives the identity of $\tan \theta \cot \theta =1$.