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Question: How do I simplify \[\dfrac{{\sin x}}{{\left( {1 - \cos x} \right)}} - \cos ecx\] \[?\]...

How do I simplify sinx(1cosx)cosecx\dfrac{{\sin x}}{{\left( {1 - \cos x} \right)}} - \cos ecx ??

Explanation

Solution

We need to simplify the trigonometric expression. We will multiply the numerator and denominator of the first term with (1+cosx)\left( {1 + \cos x} \right) so that we can get term consisting sinx\sin x in the denominator using the trigonometric identity sin2x+cos2x=1{\sin ^2}x + {\cos^2}x = 1. The second term is cosecx\cos ecx. So, we will write it as 1sinx\dfrac{1}{{\sin x}}. Now, the denominators of both the terms will become equal after cancelling common terms in numerator and denominator. Then, we can easily add up the term in numerators of both the terms.

Complete step-by-step answer:
Given to simplify sinx(1cosx)cosecx\dfrac{{\sin x}}{{\left( {1 - \cos x} \right)}} - \cos ecx
If we consider

x = \dfrac{{\sin x}}{{\left( {1 - \cos x} \right)}} \\\ y = \cos ecx \\\ $$; First we will simplify $$x$$; Multiplying $$\left( {1 + \cos x} \right)$$in both numerator and denominator , we get $$x = \dfrac{{\sin x\left( {1 + \cos x} \right)}}{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}$$ $$ \Rightarrow $$$$x = \dfrac{{\sin x\left( {1 + \cos x} \right)}}{{1 - {{\cos }^2}x}}$$ ; As we know $$\left( {a + b} \right)\left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)$$ $$ \Rightarrow x = \dfrac{{\sin x\left( {1 + \cos x} \right)}}{{{{\sin }^2}x}}$$; since $${\sin ^2}x + {\cos ^2}x = 1$$ Cancelling $$\sin x$$from numerator and denominator $$ \Rightarrow x = \dfrac{{\left( {1 + \cos x} \right)}}{{\sin x}}$$ $$ \Rightarrow x = \dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}}$$ Now we have to simplify $$y = \cos ecx$$ $$ \Rightarrow y = \dfrac{1}{{\sin x}}$$ The given question is to simplify $$\dfrac{{\sin x}}{{\left( {1 - \cos x} \right)}} - \cos ecx$$; i.e $$x - y$$; Therefore $$x - y = \dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}} - \dfrac{1}{{\sin x}}$$ Hence $$x - y = \dfrac{{\cos x}}{{\sin x}}$$; We know $$\cot x = \dfrac{{\cos x}}{{\sin x}}$$; Thus $$x - y = \cot x$$ $$\cot x$$ is the required answer . **Note:** One should have proper knowledge in trigonometry. One should learn the trigonometric formulas and identities very carefully and should try to avoid mistakes in writing them . or in applying them correctly in different kinds of sums. One must have a clear concept about the basics of trigonometry , i.e. how to derive any formula by using the right angled triangle , so that if mistakenly one forgets the formulas of trigonometry then it could be easily derived. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem.