Question

Prove that $\text{cosec }\theta +\cot \theta =\dfrac{1}{\text{cosec }\theta -\cot \theta }$ .

Answer 1

We know that the trigonometric identity $1+{{\cot }^{2}}\theta =\text{cose}{{\text{c}}^{2}}\theta$ . By using this identity, we can prove the above problem. We should rearrange the terms of this identity and then the expansion formula of the difference of squares, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to finally get the required result.

Complete step by step answer:
We know that the three most basic trigonometric identities are
$\begin{aligned} & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\\ & 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \\\ & 1+{{\cot }^{2}}\theta =\text{cose}{{\text{c}}^{2}}\theta \\\ \end{aligned}$
Let us try to prove the third identity first.
LHS = $1+{{\cot }^{2}}\theta$
We know that $\cot \theta$ is the ratio of $\cos \theta$ and $\sin \theta$. So, we can write
LHS = $1+\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }$
Taking the LCM, we get
LHS = $\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }$
But we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta$ is equal to 1. Thus, we have
LHS = $\dfrac{1}{{{\sin }^{2}}\theta }$
We know very well that $\text{cosec }\theta =\dfrac{1}{\sin \theta }$ . So, we finally have
LHS = $\text{cose}{{\text{c}}^{2}}\text{ }\theta$ = RHS
Thus, we have $1+{{\cot }^{2}}\theta =\text{cose}{{\text{c}}^{2}}\theta ...\left( i \right)$
In this problem, we will need to use equation (i). Let us rewrite this equation,
$1+{{\cot }^{2}}\theta =\text{cose}{{\text{c}}^{2}}\theta$
On rearranging the terms on both sides of the equality sign, we get
$\text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1...\left( ii \right)$
We very well know the identity for the expansion of difference of squares, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Using this identity on the LHS part of equation (ii), we get
$\left( \text{cosec }\theta +\cot \theta \right)\left( \text{cosec }\theta -\cot \theta \right)=1$
Rearranging the terms, we get
$\text{cosec }\theta +\cot \theta =\dfrac{1}{\text{cosec }\theta -\cot \theta }$.
Hence, we proved the required equation.

Note: We need to change the equation $1+{{\cot }^{2}}\theta =\text{cose}{{\text{c}}^{2}}\theta$ into the form $\text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1$. This is a relatively easy step, but we need to use this step compulsorily, to reach the required end result. We must always remember these three trigonometric identities as it forms the basis of the complete trigonometry chapter.