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# How do you simplify the expression $\dfrac{{\sin t}}{{1 - \cos t}} + \dfrac{{1 - \cos t}}{{\sin t}}$ ?

Last updated date: 17th Sep 2024
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Hint:We can see that the terms are written as in $\dfrac{a}{b} - \dfrac{b}{a}$. We will take the LCM of the terms. Then we will simplify the terms. That will lead to the standard trigonometric identity ${\sin ^2}t + {\cos ^2}t$ that is equal to 1. Then again rearranging the terms we will reach the answer. Only remember that we cannot directly get to the answer. Or we do not need to add or remove any new term in the expression. Simply LCM will help us in getting the answer.

Complete step by step answer:
Given that
$\dfrac{{\sin t}}{{1 - \cos t}} + \dfrac{{1 - \cos t}}{{\sin t}}$
Taking LCM,
$= \dfrac{{\sin t.\sin t + \left( {1 - \cos t} \right)\left( {1 - \cos t} \right)}}{{\left( {1 - \cos t} \right)\sin t}}$
Now multiplying the terms as well as the brackets in the numerator,
$= \dfrac{{{{\sin }^2}t + 1 - \cos t - \cos t + {{\cos }^2}t}}{{\left( {1 - \cos t} \right)\sin t}}$
Adding the two same terms $\cos t$,
$= \dfrac{{{{\sin }^2}t + 1 - 2\cos t + {{\cos }^2}t}}{{\left( {1 - \cos t} \right)\sin t}}$
Now we will take standard identity terms on one side,
$= \dfrac{{{{\sin }^2}t + {{\cos }^2}t + 1 - 2\cos t}}{{\left( {1 - \cos t} \right)\sin t}}$
We know that ${\sin ^2}t + {\cos ^2}t = 1$ so we will substitute the value,
$= \dfrac{{1 + 1 - 2\cos t}}{{\left( {1 - \cos t} \right)\sin t}}$
$= \dfrac{{2 - 2\cos t}}{{\left( {1 - \cos t} \right)\sin t}}$
$= \dfrac{{2\left( {1 - \cos t} \right)}}{{\left( {1 - \cos t} \right)\sin t}}$
$= \dfrac{2}{{\sin t}}$
$= 2\cos ect$
Note: Here note that the trigonometric functions are sine and cosine only used. But we need not to reshuffle any functions. We will just take the LCM in order to simplify the ratios. Also note that simplify the answer as much as we can. Such that $= \dfrac{2}{{\sin t}}$ can also be written as $= 2\cos ect$ so both steps are correct but try to write near to simplified answers.