Question

How do you simplify \[\left( {\dfrac{{\sin x}}{{1 - \cos x}}} \right) + \left( {\dfrac{{1 - \cos x}}{{\sin x}}} \right)\]?

Answer 1

Hint: To simplify this we take LCM of two fraction. We use the algebraic identity, that is \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]. We also know the Pythagoras relation between sine and cosine, that is \[{\sin ^2}x + {\cos ^2}x = 1\]. Using these identities we can solve this. We also need the reciprocal relation of six trigonometric functions.

Complete step-by-step solution:
Given, \[\left( {\dfrac{{\sin x}}{{1 - \cos x}}} \right) + \left( {\dfrac{{1 - \cos x}}{{\sin x}}} \right)\]
We know that LCM is \[\sin x(1 - \cos x)\] Simplifying we have,
\[ \Rightarrow \dfrac{{\left( {\sin x \times \sin x} \right) + \left( {\left( {1 - \cos x} \right) \times \left( {1 - \cos x} \right)} \right)}}{{\sin x\left( {1 - \cos x} \right)}}\]
Multiplying we have,
\[ \Rightarrow \dfrac{{{{\sin }^2}x + {{\left( {1 - \cos x} \right)}^2}}}{{\sin x\left( {1 - \cos x} \right)}}\]
We have identity \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\], applying this to the term \[{\left( {1 - \cos x} \right)^2}\],Where \[a = 1\] and \[b = \cos x\] then we have,
\[ \Rightarrow \dfrac{{{{\sin }^2}x + {1^2} - 2\cos x + \cos {x^2}}}{{\sin x\left( {1 - \cos x} \right)}}\]
\[ \Rightarrow \dfrac{{{{\sin }^2}x + \cos {x^2} + 1 - 2\cos x}}{{\sin x\left( {1 - \cos x} \right)}}\]
But we have the Pythagoras identity \[{\sin ^2}x + {\cos ^2}x = 1\],
\[ \Rightarrow \dfrac{{1 + 1 - 2\cos x}}{{\sin x\left( {1 - \cos x} \right)}}\]
\[ \Rightarrow \dfrac{{2 - 2\cos x}}{{\sin x\left( {1 - \cos x} \right)}}\]
Taking 2 common in the numerator we have,
\[ \Rightarrow \dfrac{{2\left( {1 - \cos x} \right)}}{{\sin x\left( {1 - \cos x} \right)}}\]
Cancelling the term we have
\[ \Rightarrow \dfrac{2}{{\sin x}}\]
But we know that the reciprocal of sine is cosecant then we have
\[ \Rightarrow 2\csc x\].
Hence, we have \[\left( {\dfrac{{\sin x}}{{1 - \cos x}}} \right) + \left( {\dfrac{{1 - \cos x}}{{\sin x}}} \right) = 2\csc x\]. This is the required answer.

Note: Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions; sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively.
Also know the relation between secant and tangent. That is
\[{\sec ^2}x - {\tan ^2}x = 1\].
We also know the relation between cosecant and cotangent. That is
\[{\csc ^2}x - {\cot ^2}x = 1\]
Remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.