Question

Consider the given trigonometric equation as $\sec \theta +\tan \theta =p$ , then find the value of $\operatorname{cosec}\theta $ is?

Answer 1

Hint: You must use the following trigonometric identities and try to find the value of \[\sec \theta \] and \[\tan \theta \] in terms of $p$ and then you can make the proper substitution to reach the final answer.
1) \[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \]
2) \[\sec \theta =\dfrac{1}{\cos \theta }\]
3) \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]
4) \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\]
To get started, use the first property and rearrange as \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\] , then apply identity \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] to get \[\left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)=1\] . We have $\sec \theta +\tan \theta =p$ from the question, substitute and then proceed further.

Complete step-by-step solution:
It is given to us that
$\sec \theta +\tan \theta =p\text{ }................\text{(1)}$
We also know that
\[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \]
The above equation can also be written as
\[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\]
Now applying the identity: \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] in the above equation, we get
\[\left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)=1\text{ }................\text{(2)}\]
Putting $\sec \theta +\tan \theta =p$ in the above equation, we get
\[p\left( \sec \theta -\tan \theta \right)=1\]
Transferring $p$ to the right hand side in the above equation, we get
\[\Rightarrow \left( \sec \theta -\tan \theta \right)=\dfrac{1}{p}\text{ }................\text{(3)}\]
Adding equation (1) and equation (3), we get
\[\begin{align}
  & 2\sec \theta =p+\dfrac{1}{p} \\
 & \Rightarrow 2\sec \theta =\dfrac{{{p}^{2}}+1}{p} \\
 & \Rightarrow \sec \theta =\dfrac{{{p}^{2}}+1}{2p} \\
\end{align}\]
Now putting \[\sec \theta =\dfrac{1}{\cos \theta }\] in the above equation, we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{\cos \theta }=\dfrac{{{p}^{2}}+1}{2p} \\
 & \Rightarrow \cos \theta =\dfrac{2p}{{{p}^{2}}+1}\text{ }................\text{(4)} \\
\end{align}\]
Now subtracting equation (3) from equation (1), we get
$\begin{align}
  & 2\tan \theta =p-\dfrac{1}{p} \\
 & \Rightarrow 2\tan \theta =\dfrac{{{p}^{2}}-1}{p} \\
 & \Rightarrow \tan \theta =\dfrac{{{p}^{2}}-1}{2p} \\
\end{align}$
Putting \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] in the above equation, we get
\[\begin{align}
  & \Rightarrow \dfrac{\sin \theta }{\cos \theta }=\dfrac{{{p}^{2}}-1}{2p} \\
 & \Rightarrow \sin \theta =\dfrac{{{p}^{2}}-1}{2p}\cos \theta \text{ }................\text{(5)} \\
\end{align}\]
Combining equation (4) with equation (5), we get
$\begin{align}
  & \sin \theta =\dfrac{{{p}^{2}}-1}{2p}\times \dfrac{2p}{{{p}^{2}}+1} \\
 & \Rightarrow \sin \theta =\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1} \\
\end{align}$
Taking the inverse of the above equation, we get
$\Rightarrow \dfrac{1}{\sin \theta }=\dfrac{{{p}^{2}}+1}{{{p}^{2}}-1}$
Putting \[\dfrac{1}{\sin \theta }=\operatorname{cosec}\theta \] in the above equation, we get
$\Rightarrow \operatorname{cosec}\theta =\dfrac{{{p}^{2}}+1}{{{p}^{2}}-1}$
So, the value of $\operatorname{cosec}\theta $ comes out to be $\dfrac{{{p}^{2}}+1}{{{p}^{2}}-1}$.

Note: Questions of trigonometry usually check your familiarity with trigonometric identities and how properly you are able to implement it. You must remember all the trigonometric identities in order to solve all the questions related to it. There is a slight possibility that you might try to break the question in terms of \[\sin \theta \] and \[\cos \theta \], but it will make things complicated for you. So always try to approach in such a way that things become easier as you go ahead and use the right identity at the right moment. In this way, you will never be stuck while solving such questions.