Question

If $\sec \theta =x+\dfrac{1}{4x}$ then prove that $\sec \theta +\tan \theta =2x\text{ or, }\dfrac{1}{2x}$.

Answer 1

Hint: Using sec$\theta$ we calculate the value of tan$\theta$ by using the identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$. After getting the value of $\tan \theta$. We can evaluate the value of sec$\theta$ + tan$\theta$. We can easily prove whether both the values given in the question are correct or not.

Complete step-by-step answer:

Given: $\sec \theta =x+\dfrac{1}{4x}$.
To prove: $\sec \theta +\tan \theta =2x\text{ or, }\dfrac{1}{2x}$.
Proof: $\sec \theta =x+\dfrac{1}{4x}$(Given in the question)
By the help of basic trigonometry identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ proceeding towards our question.
Therefore, the value of ${{\tan }^{2}}\theta$ is:
$\begin{aligned} & 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \\\ & {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1 \\\ \end{aligned}$Putting the value of $\sec \theta$ in the above equation the equation is:
$\therefore {{\tan }^{2}}\theta ={{\left( x+\dfrac{1}{4x} \right)}^{2}}-1$
Using the identity ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we can expand the above equation.
${{\tan }^{2}}\theta ={{x}^{2}}+\dfrac{1}{16{{x}^{2}}}+2\times x\times \dfrac{1}{4x}-1$
On solving the above equation, the value of ${{\tan }^{2}}\theta$ is obtained as:
$\begin{aligned} & \Rightarrow {{\tan }^{2}}\theta ={{x}^{2}}+\dfrac{1}{2}+\dfrac{1}{16{{x}^{2}}}-1 \\\ & \Rightarrow {{\tan }^{2}}\theta ={{x}^{2}}-\dfrac{1}{2}+\dfrac{1}{16{{x}^{2}}} \\\ \end{aligned}$
Here, one identity which is used can be stated as:
${{a}^{2}}-2ab+{{b}^{2}}={{(a-b)}^{2}}$
To find the value of $\tan \theta$ we solve further by using the above stated identity,
$\Rightarrow {{\tan }^{2}}\theta ={{\left( x-\dfrac{1}{4x} \right)}^{2}}$
Now taking the square root of both sides we get two values as the square produces the same result for negative and positive values.
So, the value can be expressed as $\tan \theta =x-\dfrac{1}{4x}$ or $\tan \theta =-\left( x-\dfrac{1}{4x} \right)$
Substitute the value of $\sec \theta$ and $\tan \theta$ in the given equation i.e. $\sec \theta +\tan \theta$ :
$\begin{aligned} & \sec \theta +\tan \theta =x+\dfrac{1}{4x}+x-\dfrac{1}{4x} \\\ & \sec \theta +\tan \theta =2x \\\ \end{aligned}$
$\therefore$The one value of $\sec \theta +\tan \theta =2x$ is thus proved.
Now, putting the value of $\tan \theta =-\left( x-\dfrac{1}{4x} \right)$.
$\begin{aligned} & \sec \theta +\tan \theta =x+\dfrac{1}{4x}-x+\dfrac{1}{4x} \\\ & \sec \theta +\tan \theta =\dfrac{2}{4x} \\\ \end{aligned}$
Reducing the above equation to get same expression as shown in the prove of question,
$\sec \theta +\tan \theta =\dfrac{1}{2x}$
$\therefore$ The other value of $\sec \theta +\tan \theta$ is $\dfrac{1}{2x}$.
Therefore, all the possible values of $\sec \theta +\tan \theta$ are $\dfrac{1}{2x},2x$.
Hence, we proved that $\sec \theta +\tan \theta =2x\text{ or, }\dfrac{1}{2x}$.

Note: The key step in this problem is the consideration of both the values returned by square root.
Most of the students proceed with only one value and are thus able to prove only one result due to inappropriate knowledge of square root operators. We must take both the values of $\tan \theta$.