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Question: How do you verify the identity: \({\left( {\csc x - \cot x} \right)^2} = \dfrac{{1 - \cos x}}{{1 + \...

How do you verify the identity: (csc⁡x−cot⁡x)2=1−cos⁡x1+cos⁡x{\left( {\csc x - \cot x} \right)^2} = \dfrac{{1 - \cos x}}{{1 + \cos x}} ?

Explanation

Solution

The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as cos⁡ec(x)=1sin⁡(x)\cos ec(x) = \dfrac{1}{{\sin (x)}} and cot⁡(x)=cos⁡(x)sin⁡(x)\cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}} . Basic algebraic rules and trigonometric identities are to be kept in mind while simplifying the given problem. We will first convert all the trigonometric functions into sine and cosine to deal with the problem.

Complete step by step answer:
In the given problem, we have to verify a trigonometric equality that can be further used in many questions and problems as a direct result and has wide ranging applications. For proving the desired result, we need to have a good grip over the basic trigonometric formulae and identities.Now, we need to make the left and right sides of the equation equal.
L.H.S. =(csc⁡x−cot⁡x)2 = {\left( {\csc x - \cot x} \right)^2}

We first convert all the trigonometric functions into cosine and sine to simplify the problem.Now, we know that cos⁡ec(x)=1sin⁡(x)\cos ec(x) = \dfrac{1}{{\sin (x)}} and cot⁡(x)=cos⁡(x)sin⁡(x)\cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}}. So, we get,
⇒(1sin⁡x−cos⁡xsin⁡x)2\Rightarrow {\left( {\dfrac{1}{{\sin x}} - \dfrac{{\cos x}}{{\sin x}}} \right)^2}
Since the denominator of both the terms is equal, we get,
⇒(1−cos⁡xsin⁡x)2\Rightarrow {\left( {\dfrac{{1 - \cos x}}{{\sin x}}} \right)^2}
Simplifying the expression, we get,
⇒(1−cos⁡x)2sin⁡2x\Rightarrow \dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{{{\sin }^2}x}}
Now, we know the trigonometric identity sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1. So, we get,
⇒(1−cos⁡x)21−cos⁡2x\Rightarrow \dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{1 - {{\cos }^2}x}}

Factoring the denominator using the algebraic identity a2−b2=(a−b)(a+b){a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right) , we get,
⇒(1−cos⁡x)(1−cos⁡x)(1−cos⁡x)(1+cos⁡x)\Rightarrow \dfrac{{\left( {1 - \cos x} \right)\left( {1 - \cos x} \right)}}{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}
Cancelling the common factors in numerator and denominator, we get,
⇒(1−cos⁡x)(1+cos⁡x)\Rightarrow \dfrac{{\left( {1 - \cos x} \right)}}{{\left( {1 + \cos x} \right)}}
Also, R.H.S. =(1−cos⁡x)(1+cos⁡x) = \dfrac{{\left( {1 - \cos x} \right)}}{{\left( {1 + \cos x} \right)}}
As the left side of the equation is equal to the right side of the equation, we have,
∴(csc⁡x−cot⁡x)2=1−cos⁡x1+cos⁡x\therefore {\left( {\csc x - \cot x} \right)^2} = \dfrac{{1 - \cos x}}{{1 + \cos x}}
Hence, verified.

Note: For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. We must know the algebraic identities such as a2−b2=(a−b)(a+b){a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right) for simplification of such problems.