Question

Prove the following trigonometric expression:
$\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta $

Answer 1

Hint: In trigonometry, we have some formulas that give us the relation between tan$\theta $ and cot$\theta $. One of these formulas is $\cot \theta =\dfrac{1}{\tan \theta }$. Also, we can write the trigonometric function tan$\theta $ in the form of sin$\theta $ and cos $\theta $ using the formula $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$. Also, we can write $\dfrac{1}{\sin \theta }=\csc \theta $ and $\dfrac{1}{\cos \theta }=\sec \theta $. Also, we have an identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Using these formulas, we can solve this question.

Complete step-by-step answer:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.

In trigonometry, we have a large number of formulas among which, some of them will be used to solve this question. These formulas are listed below:
$\cot \theta =\dfrac{1}{\tan \theta }$ . . . . . . . . . . . . . . . . (1)
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ . . . . . . . . . . . . . . . . . . (2)
$\dfrac{1}{\sin \theta }=\csc \theta $. . . . . . . . . . . . . . . . . . (3)
$\dfrac{1}{\cos \theta }=\sec \theta $ . . . . . . . . . . . . . . . . . . . (4)
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ . . . . . . . . . . . . . . . . (5)

In this question, we are required to prove,
$\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta $
Let us start with the left side of the above equation.
$\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }$
Using formula (1), we get,
$\begin{align}
  & \dfrac{\tan \theta }{1-\dfrac{1}{\tan \theta }}+\dfrac{\dfrac{1}{\tan \theta }}{1-\tan \theta } \\
 & \Rightarrow \dfrac{\tan \theta }{\dfrac{\tan \theta -1}{\tan \theta }}+\dfrac{\dfrac{1}{\tan \theta }}{1-\tan \theta } \\
 & \Rightarrow \dfrac{{{\tan }^{2}}\theta }{\tan \theta -1}-\dfrac{\dfrac{1}{\tan \theta }}{\tan \theta -1} \\
 & \Rightarrow \dfrac{{{\tan }^{2}}\theta -\dfrac{1}{\tan \theta }}{\tan \theta -1} \\
 & \Rightarrow \dfrac{\dfrac{{{\tan }^{3}}\theta -1}{\tan \theta }}{\tan \theta -1} \\
 & \Rightarrow \dfrac{1}{\tan \theta }\left( \dfrac{{{\tan }^{3}}\theta -1}{\tan \theta -1} \right) \\
\end{align}$

We have a formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. Using this formula, we can write ${{\tan }^{3}}\theta -1=\left( \tan \theta -1 \right)\left( {{\tan }^{2}}\theta +\tan \theta +1 \right)$.
$\begin{align}
  & \Rightarrow \dfrac{1}{\tan \theta }\left( \dfrac{\left( \tan \theta -1 \right)\left( {{\tan }^{2}}\theta +\tan \theta +1 \right)}{\tan \theta -1} \right) \\
 & \Rightarrow \dfrac{1}{\tan \theta }\left( {{\tan }^{2}}\theta +\tan \theta +1 \right) \\
 & \Rightarrow \dfrac{{{\tan }^{2}}\theta +\tan \theta +1}{\tan \theta } \\
 & \Rightarrow \dfrac{{{\tan }^{2}}\theta }{\tan \theta }+\dfrac{\tan \theta }{\tan \theta }+\dfrac{1}{\tan \theta } \\
 & \Rightarrow 1+\tan \theta +\dfrac{1}{\tan \theta } \\
\end{align}$

Using formula (2), we get,
$\begin{align}
  & 1+\dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\sin \theta } \\
 & \Rightarrow 1+\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta } \\
\end{align}$

Using formula (5), we get,
$\begin{align}
  & 1+\dfrac{1}{\sin \theta \cos \theta } \\
 & \Rightarrow 1+\dfrac{1}{\sin \theta }.\dfrac{1}{\cos \theta } \\
\end{align}$
Using formula (3) and (4), we get,
$1+\sec \theta \cos \theta $
The above equation is equal to the right side of the equation given in the question.

Hence, we have proved $\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta $.

Note: In this type of questions, the only possible mistake is when one applies an incorrect trigonometric formula. So, one must remember all the formulas thoroughly in order to obtain the correct answer in such types of questions.