Question

If $\csc \theta +\cot \theta =p$ then prove that $\cos \theta =\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}$

Answer 1

Hint: We have been given the value of p so we will start solving the question from RHS and then show that it is equal to LHS using the information given in question and from that we will find the value of ${{p}^{2}}$ and then we substitute that value of ${{p}^{2}}$ in $\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}$ and then by using some trigonometric formula we will show that it is equal to $\cos \theta $ .

Complete step-by-step solution -
Let’s start our solution.
We have been given $\csc \theta +\cot \theta =p$
Now taking square on both the sides we get,
${{p}^{2}}={{\left( \csc \theta +\cot \theta \right)}^{2}}$
Now using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ we get,
${{p}^{2}}={{\csc }^{2}}\theta +{{\cot }^{2}}\theta +2\csc \theta \cot \theta $
Now for ${{p}^{2}}-1$, we will use ${{\csc }^{2}}\theta =1+{{\cot }^{2}}\theta $ and we get ${{p}^{2}}-1$ as,
$\begin{align}
  & ={{\csc }^{2}}\theta +{{\cot }^{2}}\theta +2\csc \theta \cot \theta -1 \\
 & =1+{{\cot }^{2}}\theta +{{\cot }^{2}}\theta +2\csc \theta \cot \theta -1 \\
 & =2{{\cot }^{2}}\theta +2\csc \theta \cot \theta ..............(1) \\
\end{align}$
Now for ${{p}^{2}}+1$, we will use ${{\cot }^{2}}\theta ={{\csc }^{2}}\theta -1$ and we get ${{p}^{2}}+1$ as,
$\begin{align}
  & ={{\csc }^{2}}\theta +{{\cot }^{2}}\theta +2\csc \theta \cot \theta +1 \\
 & ={{\csc }^{2}}\theta +{{\csc }^{2}}\theta -1+2\csc \theta \cot \theta +1 \\
 & =2{{\csc }^{2}}\theta +2\csc \theta \cot \theta ............(2) \\
\end{align}$
Now for $\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}$, substituting the value of ${{p}^{2}}-1$ and ${{p}^{2}}+1$ from (1) and (2) we get,
$\begin{align}
  & =\dfrac{2{{\cot }^{2}}\theta +2\csc \theta \cot \theta }{2{{\csc }^{2}}\theta +2\csc \theta \cot \theta } \\
 & =\dfrac{2\cot \theta \left( \cot \theta +\csc \theta \right)}{2\csc \theta \left( \cot \theta +\csc \theta \right)} \\
 & =\dfrac{\cot \theta }{\csc \theta } \\
\end{align}$
Now we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and $\csc \theta =\dfrac{1}{\sin \theta }$ using these formula we get,
$\begin{align}
  & =\dfrac{\dfrac{\cos \theta }{\sin \theta }}{\dfrac{1}{\sin \theta }} \\
 & =\cos \theta \\
\end{align}$
Hence we have proved that $\cos \theta =\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}$.

Note: The formula that we have used $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and $\csc \theta =\dfrac{1}{\sin \theta }$ is very important and must be kept in mind. The most important step that we have used here was in $\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}$ , for ${{p}^{2}}-1$ we have used ${{\csc }^{2}}\theta =1+{{\cot }^{2}}\theta $ because to cancel the number 1, and for the same reason we have used ${{\cot }^{2}}\theta ={{\csc }^{2}}\theta -1$ for ${{p}^{2}}+1$ . So, this point must be clear why we are using the same formula in different ways to get the answer.