Question

If $\sec A+\tan A=p$ then find the value of $\operatorname{cosec}A$.

Answer 1

Hint: As the given expression contains trigonometric functions, so we will use trigonometric identities and formulas to solve the given question. We will use following trigonometric identity in order to solve the question:
$1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $
Also we will use relation between the trigonometric functions as follows:
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$

Complete step by step answer:
We have been given that $\sec A+\tan A=p$.
We have to find the value of $\operatorname{cosec}A$.
Now, we have $\sec A+\tan A=p........(i)$
Now, we know that $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $.
Now, we can also rewrite the above equation as
$\Rightarrow {{\sec }^{2}}A-{{\tan }^{2}}A=1$
Now, again simplifying the above obtained equation we will get
\[\Rightarrow \left( \sec A+\tan A \right)\left( \sec A-\tan A \right)=1\]
Now, substituting the value from equation (i) we will get
\[\begin{align}
  & \Rightarrow p\left( \sec A-\tan A \right)=1 \\
 & \Rightarrow \left( \sec A-\tan A \right)=\dfrac{1}{p}..............(ii) \\
\end{align}\]
Now, adding equation (i) and (ii) we will get
\[\Rightarrow \left( \sec A-\tan A \right)+\left( \sec A+\tan A \right)=\dfrac{1}{p}+p\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow 2\sec A=\dfrac{1+{{p}^{2}}}{p} \\
 & \Rightarrow \sec A=\dfrac{1+{{p}^{2}}}{2p} \\
\end{align}\]
Now, subtracting equation (ii) from (i) we will get
\[\Rightarrow \left( \sec A-\tan A \right)-\left( \sec A+\tan A \right)=p-\dfrac{1}{p}\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow 2\tan A=\dfrac{{{p}^{2}}-1}{p} \\
 & \Rightarrow \tan A=\dfrac{{{p}^{2}}-1}{2p} \\
\end{align}\]
Also we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ or $\tan \theta =\dfrac{\sec \theta }{\operatorname{cosec}\theta }$
Now, substituting the values we will get
\[\begin{align}
  & \Rightarrow \operatorname{cosec}A=\dfrac{\sec A}{\tan A} \\
 & \Rightarrow \operatorname{cosec}A=\dfrac{\left( \dfrac{1+{{p}^{2}}}{2p} \right)}{\left( \dfrac{{{p}^{2}}-1}{2p} \right)} \\
\end{align}\]
Now, simplifying the above obtained equation we will get
$\Rightarrow \operatorname{cosec}A=\dfrac{1+{{p}^{2}}}{{{p}^{2}}-1}$
Hence above is the required value of $\operatorname{cosec}A$.

Note: Alternatively we can solve the given question by using the properties of the right triangle and using Pythagoras theorem. As we know, trigonometric ratios are defined using the sides of a right triangle as hypotenuse, base and perpendicular. So we will use the properties of trigonometric functions related to solving the given question.