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Question: How do you verify \(\dfrac{{\sin x}}{{1 - \cos x}} + \dfrac{{1 - \cos x}}{{\sin x}} = 2\csc x\)?...

How do you verify sin⁡x1−cos⁡x+1−cos⁡xsin⁡x=2csc⁡x\dfrac{{\sin x}}{{1 - \cos x}} + \dfrac{{1 - \cos x}}{{\sin x}} = 2\csc x?

Explanation

Solution

To solve this question, first we have to apply the rule of addition of two fractions by taking LCM. After that we will simplify the terms. During this, we will also use some trigonometric formulas as well as the formula for expansion of a perfect square, and then prove the given equation.

Formulas used:
ab+cd=ad+bcbd\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{{ad + bc}}{{bd}}
(m−n)2=m2−2mn+n2{\left( {m - n} \right)^2} = {m^2} - 2mn + {n^2}
sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
1sin⁡x=csc⁡x\dfrac{1}{{\sin x}} = \csc x

Complete step by step answer:
We are given LHS
=sin⁡x1−cos⁡x+1−cos⁡xsin⁡x= \dfrac{{\sin x}}{{1 - \cos x}} + \dfrac{{1 - \cos x}}{{\sin x}}
We will now apply the LAM method to add these terms.
We know that, ab+cd=ad+bcbd\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{{ad + bc}}{{bd}}, where, bdbdis the LCM of bband
dd.
Here, we can take a=d=sin⁡xa = d = \sin x and b=c=1−cos⁡xb = c = 1 - \cos x.
Therefore, our LHS becomes
=(sin⁡x)2+(1−cos⁡x)2sin⁡x(1−cos⁡x)= \dfrac{{{{\left( {\sin x} \right)}^2} + {{\left( {1 - \cos x} \right)}^2}}}{{\sin x\left( {1 - \cos x} \right)}}
Now, we know the formula (m−n)2=m2−2mn+n2{\left( {m - n} \right)^2} = {m^2} - 2mn + {n^2}
Here, we have m=1m = 1 and n=cos⁡xn = \cos x
⇒(1−cos⁡x)2=1−2cos⁡x+cos⁡2x\Rightarrow {\left( {1 - \cos x} \right)^2} = 1 - 2\cos x + {\cos ^2}x
⇒LHS=sin⁡2x+1−2cos⁡x+cos⁡2xsin⁡x(1−cos⁡x)\Rightarrow LHS = \dfrac{{{{\sin }^2}x + 1 - 2\cos x + {{\cos }^2}x}}{{\sin x\left( {1 - \cos x} \right)}}
We know that sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
⇒LHS=2−2cos⁡xsin⁡x(1−cos⁡x)\Rightarrow LHS = \dfrac{{2 - 2\cos x}}{{\sin x\left( {1 - \cos x} \right)}}
Now we will take 2 common from the nominator.
⇒LHS=2(1−cos⁡x)sin⁡x(1−cos⁡x)\Rightarrow LHS = \dfrac{{2\left( {1 - \cos x} \right)}}{{\sin x\left( {1 - \cos x} \right)}}

Here, we have similar term(1−cos⁡x)\left( {1 - \cos x} \right) in both numerator and denominator. And when there are similar terms in both numerator and denominator, they will get cancelled out.
Thus, the term
(1−cos⁡x)\left( {1 - \cos x} \right)will be cancelled out.
⇒LHS=2sin⁡x\Rightarrow LHS = \dfrac{2}{{\sin x}}
We know that the reciprocal of sin⁡x\sin x is csc⁡x\csc x which means that 1sin⁡x=csc⁡x\dfrac{1}{{\sin x}} = \csc x.
⇒LHS=2csc⁡x\Rightarrow LHS = 2\csc x
⇒LHS=RHS\Rightarrow LHS = RHS
Hence, it is proved that sin⁡x1−cos⁡x+1−cos⁡xsin⁡x=2csc⁡x\dfrac{{\sin x}}{{1 - \cos x}} + \dfrac{{1 - \cos x}}{{\sin x}} = 2\csc x.

Note:
When we are solving this type of question, we need to keep in mind to put signs properly. Because a slight mistake in sign can lead us to the wrong answer or in this case, we cannot prove the left hand side is equal to the right hand side.
Another important thing is to know basic relationships between trigonometric functions. This will give us the speed for our calculations. These basic rules involve the relations among cosine, sine and tangent functions as well as the inverse functions of all these three functions.