Question

Prove that $\dfrac{\sin \theta }{1-\cos \theta }+\dfrac{\tan \theta }{1+\cos \theta }=\sec \theta \times \text{cosec}\theta +\cot \theta$

Answer 1

In this type of question we have to use the concept of trigonometry and formulae related to trigonometry. Here first consider the LHS then by performing cross multiplication we can simplify it further. Thus we can get the whole expression in terms of $\sin \theta$ and $\cos \theta$, by using $\cot \theta =\dfrac{\cos \theta }{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta },\text{cosec}\theta =\dfrac{1}{\sin \theta }$ we can obtain the required result.

Complete step-by-step solution:
Now, here we have to prove that, $\dfrac{\sin \theta }{1-\cos \theta }+\dfrac{\tan \theta }{1+\cos \theta }=\sec \theta \times \text{cosec}\theta +\cot \theta$
To prove this let us take the LHS of the above equation,
$\Rightarrow LHS=\dfrac{\sin \theta }{1-\cos \theta }+\dfrac{\tan \theta }{1+\cos \theta }$
By performing cross multiplication we can write,
$\Rightarrow LHS=\dfrac{\sin \theta \left( 1+\cos \theta \right)+\tan \theta \left( 1-\cos \theta \right)}{\left( 1-\cos \theta \right)\left( 1+\cos \theta \right)}$
As we know that, $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ by using this in the above expression we get,
$\Rightarrow LHS=\dfrac{\sin \theta +\sin \theta \cos \theta +\tan \theta -\tan \theta \cos \theta }{1-{{\cos }^{2}}\theta }$
Also by using, $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta ,\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ in the above expression we can write,
$\Rightarrow LHS=\dfrac{\sin \theta +\sin \theta \cos \theta +\dfrac{\sin \theta }{\cos \theta }-\sin \theta }{{{\sin }^{2}}\theta }$
Again by performing cross multiplication and simplifying the expression further we get,
$\Rightarrow LHS=\dfrac{\sin \theta {{\cos }^{2}}\theta +\sin \theta }{{{\sin }^{2}}\theta \cos \theta }$
Now, by taking $\sin \theta$ as a common term from the numerator we get,

& \Rightarrow LHS=\dfrac{\sin \theta \left( {{\cos }^{2}}\theta +1 \right)}{{{\sin }^{2}}\theta \cos \theta } \\\ & \Rightarrow LHS=\dfrac{{{\cos }^{2}}\theta +1}{\sin \theta \cos \theta } \\\ \end{aligned}$$ Now, by separating the denominator over the addition present in the numerator we can write, $$\begin{aligned} & \Rightarrow LHS=\dfrac{{{\cos }^{2}}\theta }{\sin \theta \cos \theta }+\dfrac{1}{\sin \theta \cos \theta } \\\ & \Rightarrow LHS=\dfrac{\cos \theta }{\sin \theta }+\dfrac{1}{\cos \theta }\centerdot \dfrac{1}{\sin \theta } \\\ \end{aligned}$$ As we know that, $$\cot \theta =\dfrac{\cos \theta }{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta },\text{cosec}\theta =\dfrac{1}{\sin \theta }$$ by using this in above expression we get, $$\Rightarrow LHS=\cot \theta +\sec \theta \text{cosec}\theta $$ By rearranging the terms we can also write the above expression as, $$\begin{aligned} & \Rightarrow LHS=\sec \theta \times \text{cosec}\theta +\cot \theta \\\ & \Rightarrow LHS=RHS \\\ \end{aligned}$$ Hence we have proved, $$\dfrac{\sin \theta }{1-\cos \theta }+\dfrac{\tan \theta }{1+\cos \theta }=\sec \theta \times \text{cosec}\theta +\cot \theta $$ **Note:** In this type of question students have to use multiple trigonometric ratios. Students have to note that it is always preferred to convert the whole expression in terms of $$\sin \theta $$ and $$\cos \theta $$. Also , students need to remember the identities such as $${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$$ not only this but also its various forms like $${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $$ or $${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $$ also. Students should not get confused with multiple formulas.