Question

Prove the following trigonometric expression:
$\cot \theta - \tan \theta = \dfrac{{2{{\cos }^2}\theta - 1}}{{\sin \theta \cos \theta }}$

Answer 1

Apply the formulas $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ on the left hand side and then cross-multiply the result to convert it in a single fraction. Then use another trigonometric formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$ to bring it in the form of the right hand side.

Complete step-by-step answer:
According to the question, we have to prove the given trigonometric equation:
$\cot \theta - \tan \theta = \dfrac{{2{{\cos }^2}\theta - 1}}{{\sin \theta \cos \theta }}{\text{ }}.....{\text{(1)}}$
We will start with the left hand side and prove that it is equal to the right hand side.
So from the above equation, we have:
$\Rightarrow LHS = \cot \theta - \tan \theta$
From the trigonometric formulas, we know that $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$. Using this in LHS, we’ll get:
$\Rightarrow LHS = \dfrac{{\cos \theta }}{{\sin \theta }} - \dfrac{{\sin \theta }}{{\cos \theta }}$
On cross-multiplication, this will give us:
$\Rightarrow LHS = \dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{\sin \theta \cos \theta }}$
Further, we also know that ${\cos ^2}\theta$ and ${\sin ^2}\theta$ are related by the formula ${\cos ^2}\theta + {\sin ^2}\theta = 1$. On rearranging this formula, we’ll get an expression of ${\sin ^2}\theta$ in terms of ${\cos ^2}\theta$:
$\Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta$
Putting this in the simplified LHS expression, we’ll get:
$\Rightarrow LHS = \dfrac{{{{\cos }^2}\theta - \left( {1 - {{\cos }^2}\theta } \right)}}{{\sin \theta \cos \theta }}$
Simplify this even further, it will give us:

$$\Rightarrow LHS = \dfrac{{{{\cos }^2}\theta - 1 + {{\cos }^2}\theta }}{{\sin \theta \cos \theta }} \\\ \Rightarrow LHS = \dfrac{{2{{\cos }^2}\theta - 1}}{{\sin \theta \cos \theta }} \\\$$

On comparing this expression with equation (1), we can say that this is equal to the right hand side.
$LHS = RHS$
Hence the equation is proved.

Note: We can also prove the equation starting from the right hand side and concluding it with the left hand side. While doing this, we have to proceed in exactly the reverse order of what we have done above. First use the formula ${\cos ^2}\theta + {\sin ^2}\theta = 1$ and convert the numerator in ${\cos ^2}\theta - {\sin ^2}\theta$. Then separate these two terms as two fractions using denominators. Finally apply the formulas $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$.