Question

Prove the following expression $\dfrac{\sin \theta }{1-\cos \theta }+\dfrac{\tan \theta }{\cos \theta +1}=\sec \theta .\csc \theta +\cot \theta $.

Answer 1

Hint: First, take the left hand side or the LHS of the given equation and simplify it by taking the LCM $\left( 1-\cos \theta \right)\left( 1+\cos \theta \right)$. Now substitute $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and cancel the like terms. After that separate the terms to prove the desired result.

Complete step-by-step answer:
Here, we have to prove that $\dfrac{\sin \theta }{1-\cos \theta }+\dfrac{\tan \theta }{\cos \theta +1}=\sec \theta .\csc \theta +\cot \theta $. Let us consider the left hand side (LHS) of the equation given in the question,
$L=\dfrac{\sin \theta }{1-\cos \theta }+\dfrac{\tan \theta }{1+\cos \theta }$
Let us simplify the above expression by taking $\left( 1-\cos \theta \right)\left( 1+\cos \theta \right)$ as the LCM. So, we get,
$L=\dfrac{\sin \theta \left( 1+\cos \theta \right)+\tan \theta \left( 1-\cos \theta \right)}{\left( 1-\cos \theta \right)\left( 1+\cos \theta \right)}$
By simplifying the above expression, we get,
$L=\dfrac{\sin \theta +\sin \theta \cos \theta +\tan \theta -\tan \theta \cos \theta }{\left( 1-\cos \theta \right)\left( 1+\cos \theta \right)}$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. By substituting this in the denominator of the above expression, we get,
$\begin{align}
  & L=\dfrac{\sin \theta +\sin \theta \cos \theta +\tan \theta -\tan \theta \cos \theta }{{{\left( 1 \right)}^{2}}-{{\left( \cos \theta \right)}^{2}}} \\
 & \Rightarrow L=\dfrac{\sin \theta +\sin \theta \cos \theta +\tan \theta -\tan \theta \cos \theta }{1-{{\cos }^{2}}\theta } \\
\end{align}$
Now, by substituting $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ in the above expression, we get,
$L=\dfrac{\sin \theta +\sin \theta \cos \theta +\dfrac{\sin \theta }{\cos \theta }-\left( \dfrac{\sin \theta }{\cos \theta } \right)\left( \cos \theta \right)}{1-{{\cos }^{2}}\theta }$
By cancelling the like terms, we get,
$L=\dfrac{\sin \theta +\sin \theta \cos \theta +\dfrac{\sin \theta }{\cos \theta }-\sin \theta }{1-{{\cos }^{2}}\theta }$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $. By using this in the denominator of the above expression, we get,
$L=\dfrac{\sin \theta +\sin \theta \cos \theta +\dfrac{\sin \theta }{\cos \theta }-\sin \theta }{{{\sin }^{2}}\theta }$
By cancelling the like terms in the above expression, we get,
$L=\dfrac{\sin \theta \cos \theta +\dfrac{\sin \theta }{\cos \theta }}{{{\sin }^{2}}\theta }$
By separating the terms of the above expression, we get,
$L=\dfrac{\sin \theta \cos \theta }{\left( {{\sin }^{2}}\theta \right)}+\dfrac{\sin \theta }{\cos \theta .\left( {{\sin }^{2}}\theta \right)}$
By cancelling the like terms from each term of the above expression, we get,
$L=\dfrac{\cos \theta }{\sin \theta }+\dfrac{1}{\cos \theta .\sin \theta }$
We know that $\dfrac{\cos \theta }{\sin \theta }=\cot \theta $, so, by using this in the above expression, we get,
$L=\cot \theta +\dfrac{1}{\sin \theta }.\dfrac{1}{\cos \theta }$
We also know that $\dfrac{1}{\sin \theta }=\csc \theta $ and $\dfrac{1}{\cos \theta }=\sec \theta $. By using this in the above expression, we get,
$L=\cot \theta +\csc \theta .\sec \theta $, which is equal to the right hand side or RHS.
Hence, we have proved that LHS = RHS or $\dfrac{\sin \theta }{1-\cos \theta }+\dfrac{\tan \theta }{\cos \theta +1}=\sec \theta .\csc \theta +\cot \theta $.

Note: In these types of questions, it is always advisable to simplify the expression and convert the whole expression in terms of $\sin \theta $ and $\cos \theta $. Also, in trigonometry it is very important to remember the formulas, so the students must make a list of the trigonometric formulas and make use of it while solving questions related to trigonometry. Also, check the equation twice after writing it to avoid any mistakes.