Question

# Prove that: $\frac{{\sin \theta }}{{1 - \cos \theta }} = cosec\theta + \cot \theta$

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Hint: - Taking conjugate of denominator.
Given,
L.H.S $\begin{gathered} = \frac{{\sin \theta }}{{1 - \cos \theta }} \\ \\ \end{gathered}$
Multiply and Divide by $\left( {1 + \cos \theta } \right)$ , we get
$\begin{gathered} = \frac{{\sin \theta }}{{1 - \cos \theta }} \times \frac{{1 + \cos \theta }}{{1 + \cos \theta }} \\ = \frac{{\sin \theta \left( {1 + \cos \theta } \right)}}{{\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)}} \\ = \frac{{\sin \theta \left( {1 + \cos \theta } \right)}}{{{{\left( 1 \right)}^2} - {{\left( {\cos \theta } \right)}^2}}} \\ = \frac{{\sin \theta \left( {1 + \cos \theta } \right)}}{{1 - {{\cos }^2}\theta }} \\ \end{gathered}$
We know that, ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Or, ${\sin ^2}\theta = 1 - {\cos ^2}\theta$
Replace $\left( {1 - {{\cos }^2}\theta } \right)$ by ${\sin ^2}\theta$ , we get
L.H.S $\begin{gathered} = \frac{{\sin \theta \left( {1 + \cos \theta } \right)}}{{{{\sin }^2}\theta }} \\ = \frac{{\sin \theta \left( {1 + \cos \theta } \right)}}{{\sin \theta \times \sin \theta }} \\ \end{gathered}$
Cancel out $\sin \theta$ in numerator by $\sin \theta$ in denominator, we get
$\begin{gathered} = \frac{{1 + \cos \theta }}{{\sin \theta }} \\ = \frac{1}{{\sin \theta }} + \frac{{\cos \theta }}{{\sin \theta }} \\ \end{gathered}$
Now we can written $\frac{1}{{\sin \theta }} = \cos ec\theta$ and $\frac{{\cos \theta }}{{\sin \theta }} = \cot \theta$ , we get
L.H.S $= \cos ec\theta + \cot \theta$- (1)
Since, given R.H.S$= \cos ec\theta + \cot \theta$ - (2)
By seeing equation (1) and (2) we can tell that
L.H.S R.H.S
Hence, it proved.
Note: - These types of questions are also solve by taking R.H.S (Right Hand Side), solve it to prove L.H.S (Left Hand Side). During solving trigonometry proving we should always have basic trigonometry identities in our mind.