Question

How do you simplify $\left( \cot x+\csc x \right)\left( \cot x-\csc x \right)$?

Answer 1

Hint: In this problem we need to simplify the given equation i.e., we need to calculate the value of the given equation. In the given equation we can observe that the trigonometric functions $\cot $, $\csc $ are in addition and subtraction. So, we will first consider the term $\cot x+\csc x$ and we will simplify this term by using the basic trigonometric definitions we have. Now we will get the result in terms of $\sin x$, $\cos x$. Now we will follow the same procedure for simplifying the term $\cot x-\csc x$. Now we will multiply the both terms and use algebraic formulas and trigonometric identity to get the result.

Complete step by step answer:
Given that, $\left( \cot x+\csc x \right)\left( \cot x-\csc x \right)$.
Considering the term $\cot x+\csc x$ separately. Now we have $\cot x=\dfrac{\cos x}{\sin x}$, $\csc x=\dfrac{1}{\sin x}$. Substituting these values in the above term, then we will get
$\cot x+\csc x=\dfrac{\cos x}{\sin x}+\dfrac{1}{\sin x}$
Simplifying the above equation by taking LCM, then we will get
$\Rightarrow \cot x+\csc x=\dfrac{\cos x+1}{\sin x}$
Similarly, we will get the value of $\cot x-\csc x$ as
$\cot x-\csc x=\dfrac{\cos x-1}{\sin x}$
Now the value of $\left( \cot x+\csc x \right)\left( \cot x-\csc x \right)$ will be
$\left( \cot x+\csc x \right)\left( \cot x-\csc x \right)=\left( \dfrac{\cos x+1}{\sin x} \right)\left( \dfrac{\cos x-1}{\sin x} \right)$
Using the algebraic formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, then we will get
$\Rightarrow \left( \cot x+\csc x \right)\left( \cot x-\csc x \right)=\dfrac{{{\cos }^{2}}x-1}{{{\sin }^{2}}x}$
From the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, the value of ${{\cos }^{2}}x-1=-{{\sin }^{2}}x$. Substituting this value in the above equation, then we will get
$\begin{align}
  & \Rightarrow \left( \cot x+\csc x \right)\left( \cot x-\csc x \right)=\dfrac{-{{\sin }^{2}}x}{{{\sin }^{2}}x} \\
 & \therefore \left( \cot x+\csc x \right)\left( \cot x-\csc x \right)=-1 \\
\end{align}$

Note: In the above problem we have used the algebraic formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ after converting each term into function of $\sin x$, $\cos x$. We can also use the same formula without converting each term, then we will get
$\Rightarrow \left( \cot x+\csc x \right)\left( \cot x-\csc x \right)={{\cot }^{2}}x-{{\csc }^{2}}x$
We have the trigonometric identity ${{\csc }^{2}}x-{{\cot }^{2}}x=1$. From this identity we can write
$\Rightarrow \left( \cot x+\csc x \right)\left( \cot x-\csc x \right)=-1$
From both the methods we got the same result.