Question

If a trigonometric equation is given as $\sec \theta +\tan \theta =p$, then find the value of $\cos ec\theta $.

Answer 1

Hint: To solve this problem, we should be aware of the basic formulas of trigonometric angles. Further, we would then use the property that ${{\sec }^{2}}x-{{\tan }^{2}}x=1$ to simplify the problem in hand.

Complete step-by-step solution -
To solve the above problem, we first start by assuming $\sec \theta -\tan \theta =a$. We will then proceed by simplifying the equation. We get,
$\sec \theta +\tan \theta =p$ -- (1)
Also, we have,
$\sec \theta -\tan \theta =a$ -- (2)
We multiply (1) and (2), we get,
$\begin{align}
  & (\sec \theta -\tan \theta )(\sec \theta +\tan \theta )=ap \\
 & {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =ap \\
\end{align}$
Now, we know that ${{\sec }^{2}}x-{{\tan }^{2}}x=1$, thus, we have,
ap = 1
Thus, a = $\dfrac{1}{p}$
Thus, $\sec \theta -\tan \theta =\dfrac{1}{p}$ -- (3)
Now, we add (1) and (3), we get,
2 $\sec \theta $ = p + $\dfrac{1}{p}$
$\sec \theta $ = $\dfrac{1}{2}\left( p+\dfrac{1}{p} \right)$ -- (A)
Now, putting this value in equation (1), we get,
$\sec \theta +\tan \theta =p$
$\dfrac{1}{2}\left( p+\dfrac{1}{p} \right)+\tan \theta =p$
$\tan \theta =\dfrac{1}{2}\left( p-\dfrac{1}{p} \right)$ -- (B)
From (A), we have,
$\sec \theta $ = $\dfrac{1}{2}\left( p+\dfrac{1}{p} \right)$
$\cos \theta =\dfrac{1}{\sec \theta }$
Thus, $\cos \theta =\dfrac{2p}{{{p}^{2}}+1}$ -- (C)
Also, from (B), we have,
$\tan \theta =\dfrac{1}{2}\left( p-\dfrac{1}{p} \right)$
$\dfrac{\sin \theta }{\cos \theta }=\dfrac{1}{2}\left( p-\dfrac{1}{p} \right)$
Substituting the value from (C), we get,
$\begin{align}
  & \dfrac{\sin \theta }{\dfrac{2p}{{{p}^{2}}+1}}=\dfrac{1}{2}\left( p-\dfrac{1}{p} \right) \\
 & \sin \theta =\dfrac{2p}{{{p}^{2}}+1}\left( \dfrac{1}{2} \right)\left( p-\dfrac{1}{p} \right) \\
\end{align}$
Thus, we have,
$\begin{align}
  & \sin \theta =\dfrac{p}{{{p}^{2}}+1}\left( p-\dfrac{1}{p} \right) \\
 & \sin \theta =\dfrac{p}{{{p}^{2}}+1}\left( \dfrac{{{p}^{2}}-1}{p} \right) \\
 & \sin \theta =\left( \dfrac{{{p}^{2}}-1}{{{p}^{2}}+1} \right) \\
\end{align}$
Now, we know that $\cos ec\theta =\dfrac{1}{\sin \theta }$. Thus, we have,
$\begin{align}
  & \sin \theta =\left( \dfrac{{{p}^{2}}-1}{{{p}^{2}}+1} \right) \\
 & \cos ec\theta =\left( \dfrac{{{p}^{2}}+1}{{{p}^{2}}-1} \right) \\
\end{align}$
Hence, the correct answer is $\left( \dfrac{{{p}^{2}}+1}{{{p}^{2}}-1} \right)$.

Note: One can get stuck in these problems by converting secx and tanx into respective sine and cosine functions. Doing this, leads to an unnecessary and elaborate equation, which cannot be solved easily. As a general thumb rule, when given problems of trigonometry in which secx and tanx appear together in an equation, one should always think about utilizing ${{\sec }^{2}}x-{{\tan }^{2}}x=1$ property first. In most cases, this greatly simplifies the equation and easily gives the required solution. If the equation is still unsolvable, one can then convert into sinx and cosx and then proceed to simplify the trigonometric equation.