Question

If we have a trigonometric expression $\sec \theta +\tan \theta =x$ , then which of the following is the value of $\sec \theta $ .
A. $\dfrac{{{x}^{2}}+1}{x}$
B. $\dfrac{{{x}^{2}}+1}{2x}$
C. $\dfrac{{{x}^{2}}-1}{2x}$
D. $\dfrac{{{x}^{2}}-1}{x}$

Answer 1

Hint: To find the value of $\sec \theta $ , we will consider $\sec \theta +\tan \theta =x...(i)$. Multiply and divide (i) with $\sec \theta -\tan \theta $ . Using ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ and solving this further, we will get $\sec \theta -\tan \theta =\dfrac{1}{x}...(ii)$ . When we add equations (i) and (ii) and solve, the value of $\sec \theta $ can be obtained.

Complete step-by-step solution
It is given that $\sec \theta +\tan \theta =x$ . We have to find the value of $\sec \theta $ .
Let us consider $\sec \theta +\tan \theta =x...(i)$
We have to multiply and divide equation (i) with $\sec \theta -\tan \theta $ . We will get
$\dfrac{\left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)}{\sec \theta -\tan \theta }=x$
Now, let us expand the numerator. We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Hence,
$\dfrac{{{\sec }^{2}}\theta -{{\tan }^{2}}\theta }{\sec \theta -\tan \theta }=x$
We know that ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ . Hence, the above equation becomes
$\dfrac{1}{\sec \theta -\tan \theta }=x$
This can be written as
$\sec \theta -\tan \theta =\dfrac{1}{x}...(ii)$
Now, let us add equations (i) and (ii). This is shown below.
$\begin{align}
  & \sec \theta +\tan \theta +\sec \theta -\tan \theta =x+\dfrac{1}{x} \\
 & \Rightarrow 2\sec \theta =x+\dfrac{1}{x} \\
\end{align}$
When we solve the RHS, we will get
$2\sec \theta =\dfrac{{{x}^{2}}+1}{x}$
Now, we can find the value of $\sec \theta $ by taking 2 to the RHS. We will get
$\sec \theta =\dfrac{{{x}^{2}}+1}{2x}$
Hence, the correct option is B.

Note: You must know the trigonometric identities to solve this question. You may make mistake when writing the identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ as ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =-1$ . We can also solve this question in an alternate method as shown below.
We know that if $\sec \theta +\tan \theta =a$ then $\sec \theta -\tan \theta =\dfrac{1}{a}$ .
We have $\sec \theta +\tan \theta =x...(i)$ . Then,
$\sec \theta -\tan \theta =\dfrac{1}{x}...\left( ii \right)$
Now, let us add (i) and (ii). We will get
$\sec \theta +\tan \theta +\sec \theta -\tan \theta =x+\dfrac{1}{x}$
$\Rightarrow 2\sec \theta =x+\dfrac{1}{x}$
When we solve the RHS, we will get
$2\sec \theta =\dfrac{{{x}^{2}}+1}{x}$
Now, we can find the value of $\sec \theta $ by taking 2 to the RHS. We will get
$\sec \theta =\dfrac{{{x}^{2}}+1}{2x}$