Question

How do you verify the identity ${(\csc x - \cot x)^2} = \dfrac{{1 - \cos x}}{{1 + \cos x}}$?

Answer 1

The given trigonometric is ${(\csc x - \cot x)^2} = \dfrac{{1 - \cos x}}{{1 + \cos x}}$
First we let’s take RHS (Right Hand Side)
After that we simplify the trigonometric function.
Finally we get the proof of identities in the given trigonometric function.
We use the expansion: ${(a - b)^2} = {a^2} - 2ab + {b^2}$
And then we use the fundamental identities (Pythagorean identities) of trigonometry.
${\cos ^2}\theta + {\sin ^2}\theta = 1$
Finally we get the required answer.

Complete step-by-step solution:
The given trigonometric is ${(\csc x - \cot x)^2} = \dfrac{{1 - \cos x}}{{1 + \cos x}}$
Let’s take RHS (Right Hand Side)
$\Rightarrow {(\csc x - \cot x)^2}$
This in the form of ${(a - b)^2}$
We expand the equation
$\Rightarrow {(a - b)^2} = {a^2} - 2ab + {b^2}$
Hence we apply the expansion in the trigonometric function, hence we get
$\Rightarrow {(\csc x - \cot x)^2} = ({\csc ^2}x - 2\cot x\csc x + {\cot ^2}x)$
We use the formula given below,
$\Rightarrow \csc x = \dfrac{1}{{\sin x}},\cot x = \dfrac{{\cos x}}{{\sin x}}$
This formula apply in the expansion of ${(\csc x - \cot x)^2}$, hence we get
$\Rightarrow {(\csc x - \cot x)^2} = \left( {\dfrac{1}{{{{\sin }^2}x}} - 2\dfrac{{\cos x}}{{\sin x}} \times \dfrac{1}{{\sin x}} + \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right)$
Multiply the second term $\cos x$ by $1$ in the numerator and multiply $\sin x$ by $\sin x$ in the denominator, hence we get
$\Rightarrow {(\csc x - \cot x)^2} = \left( {\dfrac{1}{{{{\sin }^2}x}} - 2\dfrac{{\cos x}}{{{{\sin }^2}x}} + \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right)$
We take the common term ${\sin ^2}x$ over the denominator, hence we get
$\Rightarrow {(\csc x - \cot x)^2} = \left( {\dfrac{{1 - 2\cos x + {{\cos }^2}x}}{{{{\sin }^2}x}}} \right)$
The numerator we can rewrite, hence we get
$\Rightarrow {(\csc x - \cot x)^2} = \left( {\dfrac{{(1 - \cos x)(1 - \cos x)}}{{{{\sin }^2}x}}} \right)$
And then we use the fundamental identities (Pythagorean identities) of trigonometry.
$\Rightarrow {\cos ^2}\theta + {\sin ^2}\theta = 1$
$\Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta$
Then we substitute$\theta = x$, ${\sin ^2}x = 1 - {\cos ^2}x$
We apply the identities in the denominator, hence we get
$\Rightarrow {(\csc x - \cot x)^2} = \left( {\dfrac{{(1 - \cos x)(1 - \cos x)}}{{1 - {{\cos }^2}x}}} \right)$
We rewrite the denominator, hence we get
$\Rightarrow {(\csc x - \cot x)^2} = \left( {\dfrac{{\not{{(1 - \cos x)}}(1 - \cos x)}}{{\not{{(1 - \cos x)}}(1 + \cos x)}}} \right)$
$\Rightarrow {(\csc x - \cot x)^2} = \left( {\dfrac{{(1 - \cos x)}}{{(1 + \cos x)}}} \right)$
Hence we verify that ${(\csc x - \cot x)^2} = \left( {\dfrac{{(1 - \cos x)}}{{(1 + \cos x)}}} \right)$.

Note: An even function is symmetric (by reflection) about the $y$-axis, i.e. $\operatorname{f} ( - x) = f(x)$
An odd function is symmetric (by ${180^ \circ }$ rotation) about the origin, i.e. $\operatorname{f} ( - x) = - f(x)$
The following shows the even trigonometric functions and odd trigonometric functions.
Even trigonometric functions and identities:
The Cosine function is even. $\cos ( - x) = \cos (x)$
The Secant function is even. $\sec ( - x) = \sec (x)$
Odd trigonometric functions and identities:
The Sine function is odd. $\sin ( - x) = - \sin (x)$
The Cosecant function is odd. $\csc ( - x) = - \csc (x)$
The Tangent function is odd. $\tan ( - x) = - \tan (x)$
The Cotangent function is odd. $\cot ( - x) = - \cot (x)$