Question

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

Answer 1

To prove the given trigonometric function, apply trigonometric identity formulas i.e., applying $\csc x$ formula in terms of sin function and cot function, as we know that sec, csc and cot are derived from primary functions of sin, cos and tan, hence using these functions we can prove that LHS = RHS of the given function.

Complete step-by-step solution:
The given function is
$\Rightarrow \left( {\csc x + \cot x} \right)\left( {\csc x - \cot x} \right) = 1$
We know that
$\Rightarrow \csc x = \dfrac{1}{{\sin x}}$ and
$\Rightarrow \cot x = \dfrac{{\cos x}}{{\sin x}}$
Therefore,
$\Rightarrow {\sin ^2}x + {\cos ^2}x = 1$
Now let us consider the LHS term i.e.,
$\Rightarrow \left( {\csc x + \cot x} \right)\left( {\csc x - \cot x} \right)$
Simplifying and combining the terms we get
$\Rightarrow {\csc ^2}x - {\cos ^2}x$ ………………. 1
As we know the formula of ${\csc ^2}x$ is
$\Rightarrow \dfrac{1}{{{{\sin }^2}x}}$, hence apply this in equation 1 as
$\Rightarrow \dfrac{1}{{{{\sin }^2}x}} - \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}$
Which implies to
$\Rightarrow \dfrac{{1 - {{\cos }^2}x}}{{{{\sin }^2}x}}$
We know that $1 - {\cos ^2}x$ is ${\sin ^2}x$, hence we get
$\Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\sin }^2}x}}$
$\Rightarrow 1 = RHS$

Therefore, hence proved $\left( {\csc x + \cot x} \right)\left( {\csc x - \cot x} \right) = 1$

Additional Information:
The three basic functions in trigonometry are sine, cosine and tangent. Based on these three functions the other three functions that are cotangent, secant and cosecant are derived. All the trigonometrical concepts are based on these functions. Hence, to understand trigonometry further we need to learn these functions and their respective formulas at first.
If θ is the angle in a right-angled triangle, then
Sin θ = $\dfrac{{perpendicular}}{{hypotenuse}}$
Cos θ = $\dfrac{{base}}{{hypotenuse}}$
Tan θ = $\dfrac{{perpendicular}}{{base}}$
The other three functions i.e., cot, sec and cosec depend on tan, cos and sin respectively.

Note: The key point to prove any trigonometric function is to note the chart of all related functions with respect to the equation, and prove all the terms by considering LHS terms and prove for RHS. And here are some of the formulas to be noted.
$1 - {\cos ^2}x$ = ${\sin ^2}x$ , ${\sin ^2}x + {\cos ^2}x = 1$,
Cot θ = $\dfrac{{\cos \theta }}{{\sin \theta }}$, Cosec θ = $\dfrac{{\sec \theta }}{{\tan \theta }}$