Question

How do you prove ${{\csc }^{2}}\left( x \right)-{{\cot }^{2}}\left( x \right)=1$?

Answer 1

To solve this question, we will need to use the trigonometric identity which states that ${{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)=1$. To prove an equation, we need to take one of the left-hand side (LHS) or right-hand side (RHS). Simplify the side and express it as the other side.

Complete step by step answer:
We are asked to prove ${{\csc }^{2}}\left( x \right)-{{\cot }^{2}}\left( x \right)=1$. Here the left-hand side is ${{\csc }^{2}}\left( x \right)-{{\cot }^{2}}\left( x \right)$, and the right-hand side is $1$. We need to choose one of the sides and simplify the side so it can be expressed as the other side.
Let’s choose the left-hand side ${{\csc }^{2}}\left( x \right)-{{\cot }^{2}}\left( x \right)$, we know that $\csc x=\dfrac{1}{\sin x}$, and $\cot x=\dfrac{\cos x}{\sin x}$. Using this in the expression, it can be expressed as
$\Rightarrow {{\left( \dfrac{1}{\sin x} \right)}^{2}}-{{\left( \dfrac{\cos x}{\sin x} \right)}^{2}}$
$\Rightarrow \dfrac{1}{{{\sin }^{2}}x}-\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}$
As the denominator for both terms in the above expression is the same, we can subtract the numerators,
$\Rightarrow \dfrac{1-{{\cos }^{2}}x}{{{\sin }^{2}}x}$
Now, we know the trigonometric identity which states that ${{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)=1$, subtracting ${{\cos }^{2}}\left( x \right)$ from both sides of this identity, we get
$\Rightarrow {{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)-{{\cos }^{2}}\left( x \right)=1-{{\cos }^{2}}\left( x \right)$
$\Rightarrow {{\sin }^{2}}\left( x \right)=1-{{\cos }^{2}}\left( x \right)$
Using this identity in the above expression of the left-hand side, we get
$\Rightarrow \dfrac{1-{{\cos }^{2}}x}{{{\sin }^{2}}x}=\dfrac{1-{{\cos }^{2}}x}{1-{{\cos }^{2}}x}$
$\Rightarrow 1$
After simplifying the left-hand side, we get the 1, as the right-hand side of the given expression is also 1. Hence, we proved that the left-hand side and right-hand sides are equal.
Hence, ${{\csc }^{2}}\left( x \right)-{{\cot }^{2}}\left( x \right)=1$ is proved.

Note: The expression given in the question is one of the trigonometric identities similar to ${{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)=1$. We can also prove that $1+{{\tan }^{2}}\left( x \right)={{\sec }^{2}}\left( x \right)$ by using a similar method to this question.
The following identities are useful while solving the question on proofs or evaluating expressions, so it should be remembered:
${{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)=1$
${{\csc }^{2}}\left( x \right)=1+{{\cot }^{2}}\left( x \right)$
$1+{{\tan }^{2}}\left( x \right)={{\sec }^{2}}\left( x \right)$