Question

Consider the condition, $\sec \theta -\tan \theta =5$, then find the value of $\theta $ .

Answer 1

Hint: Factorize ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta $ by using an algebraic identity given as ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ . Now, replace ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta $ by 1 using relation ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ in one side and put value of $\sec \theta -\tan \theta $ to other side as given in the question. Now, get the value of $\sec \theta +\tan \theta $ and solve both the equations of $\sec \theta -\tan \theta $ and $\sec \theta +\tan \theta $ to get the value of $\theta $ .

“Complete step-by-step answer:”
Here, we have
$\sec \theta -\tan \theta =5...............\left( i \right)$
So, we need to determine the value of ‘$\theta $‘ from the given relation of the equation (i).
As we know trigonometric identity related with $\sec \theta $ and $\tan \theta $ can be given as ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1.$ And we know identity of $\left( {{a}^{2}}-{{b}^{2}} \right)$ in algebraic equations can be given as (a - b) (a + b). If we compare ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta $ with ${{a}^{2}}-{{b}^{2}}$ , then we can factorize ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta $ by relation ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ as,
${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =\left( \sec \theta -\tan \theta \right)\left( \sec \theta +\tan \theta \right).........\left( ii \right)$
Now, we can put value of $\sec \theta -\tan \theta $ as ‘5’ from the question (equation(i)) and replace ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta $ by 1 using the above mentioned identity as ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1.$ Hence, we can re-write the equation (ii) as
$1=\left( \sec \theta +\tan \theta \right)5$
$\Rightarrow \sec \theta +\tan \theta =\dfrac{1}{5}.................\left( iii \right)$
Now, we can add equation (i) and (iii) to get the value of ‘$\theta $’. Hence, on adding equation (i) and (iii); we get
$\begin{align}
  & \left( \sec \theta -\tan \theta \right)+\left( \sec \theta +\tan \theta \right)=\dfrac{5}{1}+\dfrac{1}{5} \\
 & \sec \theta -\tan \theta +\sec \theta +\tan\theta =\dfrac{25+1}{5} \\
 & 2\sec \theta =\dfrac{26}{5} \\
\end{align}$
$\begin{align}
  & \Rightarrow \sec \theta =\dfrac{26}{5\times 2}=\dfrac{13}{5} \\
 & \Rightarrow \sec \theta =\dfrac{13}{5} \\
\end{align}$
We can convert $\sec \theta $ to $\cos \theta $ by using the relation $\cos \theta =\dfrac{1}{\sec \theta }$
Now, put the value of $\sec \theta $ in the above relation to get the value of $\theta $ in $\cos $ form. Hence, we get
$\begin{align}
  & \cos \theta =\dfrac{1}{\left( \dfrac{13}{5} \right)}=\dfrac{5}{13} \\
 & \cos \theta =\dfrac{5}{13} \\
\end{align}$
$\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{5}{13} \right)$
This is the required value.

Note: Another approach to solve the given question would be that we can find the value of $\theta $ by using the relations $\sec \theta =\dfrac{1}{\cos \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ . Hence, we can get equation as
$\dfrac{1}{\cos \theta }-\dfrac{\sin \theta }{\cos \theta }=5$
$1-\sin \theta =5\cos \theta $
Now, square both sides and replace \[{{\cos }^{2}}\theta \] by $1-{{\sin }^{2}}\theta $ using relation ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.$ Hence, we get
$1+{{\sin }^{2}}\theta -2\sin \theta =25\left( 1-{{\sin }^{2}}\theta \right)$
$26{{\sin }^{2}}\theta -2\sin \theta -24=0$
$13{{\sin }^{2}}\theta -\sin \theta -12=0$
Now, solve the quadratic and hence get the value of $\sin \theta $ and further change $\sin \theta $ to $\cos \theta $ as well to get the same answer as in the solution.
One can get confused with the relation ${{\sin }^{2}}\theta -{{\tan }^{2}}\theta =1$ . He or she may use ${{\sec }^{2}}\theta +{{\tan }^{2}}\theta =1$ which is wrong. So, be clear with the trigonometric identities. And writing ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta $ to $\left( \sec \theta -\tan \theta \right)\left( \sec \theta +\tan \theta \right)$ using identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ is the key point of the question and solution.