Question

If $\dfrac{\left( \sec \theta +\tan \theta \right)}{\left( \sec \theta -\tan \theta \right)}=\dfrac{209}{79}$, find the value of $\theta $.

Answer 1

Hint:Rationalize the L.H.S by multiplying it with the conjugate of denominator given by: $\dfrac{\left( \sec \theta +\tan \theta \right)}{\left( \sec \theta +\tan \theta \right)}$. Use the trigonometric identity: ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ to simplify the denominator of L.H.S. Now, convert the given trigonometric functions into its sine and cosine form by using the relations: $\sec \theta =\dfrac{1}{\cos \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$. Use the identity: ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ to get a quadratic equation in $\sin \theta $. Solve this equation to find the general solution. Use the relation: if $\sin x=a$, then its general solution is given by $x=n\pi +{{\left( -1 \right)}^{n}}{{\sin }^{-1}}\left( a \right)$, where ‘n’ is any integer. Remember that $\theta $ cannot be any odd multiple of $\dfrac{\pi }{2}$ because at that value tangent and secant are undefined.

Complete step-by-step answer:
We have been provided with the equation: $\dfrac{\left( \sec \theta +\tan \theta \right)}{\left( \sec \theta -\tan \theta \right)}=\dfrac{209}{79}$.
Rationalizing the denominator in the L.H.S, we get,
$\begin{align}
  & \dfrac{\left( \sec \theta +\tan \theta \right)}{\left( \sec \theta -\tan \theta \right)}\times \dfrac{\left( \sec \theta +\tan \theta \right)}{\left( \sec \theta +\tan \theta \right)}=\dfrac{209}{79} \\
 & \Rightarrow \dfrac{{{\left( \sec \theta +\tan \theta \right)}^{2}}}{\left( \sec \theta -\tan \theta \right)\left( \sec \theta +\tan \theta \right)}=\dfrac{209}{79} \\
 & \Rightarrow \dfrac{{{\left( \sec \theta +\tan \theta \right)}^{2}}}{\left( {{\sec }^{2}}\theta -{{\tan }^{2}}\theta \right)}=\dfrac{209}{79} \\
\end{align}$
Using the identity: ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$, we get,
${{\left( \sec \theta +\tan \theta \right)}^{2}}=\dfrac{209}{79}$
Now, using the relations: $\sec \theta =\dfrac{1}{\cos \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, we get,
$\begin{align}
  & {{\left( \dfrac{1}{\cos \theta }+\dfrac{\sin \theta }{\cos \theta } \right)}^{2}}=\dfrac{209}{79} \\
 & \Rightarrow {{\left( \dfrac{1+\sin \theta }{\cos \theta } \right)}^{2}}=\dfrac{209}{79} \\
 & \Rightarrow \left( \dfrac{1+{{\sin }^{2}}\theta +2\sin \theta }{{{\cos }^{2}}\theta } \right)=\dfrac{209}{79} \\
\end{align}$
By cross-multiplication, we have,
$79+79{{\sin }^{2}}\theta +158\sin \theta =209{{\cos }^{2}}\theta $
Using the identity: ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$, we get,
$\begin{align}
  & 79+79{{\sin }^{2}}\theta +158\sin \theta =209\left( 1-{{\sin }^{2}}\theta \right) \\
 & \Rightarrow 79+79{{\sin }^{2}}\theta +158\sin \theta =209-209{{\sin }^{2}}\theta \\
 & \Rightarrow 288{{\sin }^{2}}\theta +158\sin \theta -130=0 \\
 & \Rightarrow 2\left( 144{{\sin }^{2}}\theta +79\sin \theta -65 \right)=0 \\
 & \Rightarrow \left( 144{{\sin }^{2}}\theta +79\sin \theta -65 \right)=0 \\
\end{align}$
This is a quadratic equation in $\sin \theta $. Splitting the middle term, we get,
$\begin{align}
  & 144{{\sin }^{2}}\theta +144\sin \theta -65\sin \theta -65=0 \\
 & \Rightarrow 144\sin \theta \left( \sin \theta +1 \right)-65\left( \sin \theta +1 \right)=0 \\
 & \Rightarrow \left( 144\sin \theta -65 \right)\left( \sin \theta +1 \right)=0 \\
\end{align}$
Equating each term equal to 0, we get,
$\begin{align}
  & \left( 144\sin \theta -65 \right)=0\text{ or }\left( \sin \theta +1 \right)=0 \\
 & \Rightarrow \sin \theta =\dfrac{65}{144}\text{ or }\sin \theta =-1 \\
\end{align}$
Now, we know that the value of $\sin \theta $ is 1 or -1, when the angle is an odd multiple of $\dfrac{\pi }{2}$. But the functions: $\tan \theta $ and $\sec \theta $ are undefined for such angles. Therefore the only solution will be $\sin \theta =\dfrac{65}{144}$.
Now, we have to find a general solution.
If $\sin x=a$, then its general solution is given by $x=n\pi +{{\left( -1 \right)}^{n}}{{\sin }^{-1}}\left( a \right)$, where ‘n’ is any integer. So, the general solution of $\sin \theta =\dfrac{65}{144}$ is given as:
$\theta =n\pi +{{\left( -1 \right)}^{n}}{{\sin }^{-1}}\left( \dfrac{65}{144} \right)$, where ‘n’ is any integer.

Note: One may note that we have to find the solution according to the equation: $\dfrac{\left( \sec \theta +\tan \theta \right)}{\left( \sec \theta -\tan \theta \right)}=\dfrac{209}{79}$ and not its simplified form. So, wherever this equation is undefined, we have to reject that value from the obtained solution. That is why, $\sin \theta =-1$ is not considered because it will give such values of $\theta $, at which secant and tangent functions are undefined.