Question

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

Answer 1

Hint: We have been given $\sec \theta =x+\dfrac{1}{4x}$. So use the formula ${{\sec }^{2}}\theta -1={{\tan }^{2}}\theta $ and simplify. You will get the value of $\tan \theta $. After that add $\sec \theta $ and $\tan \theta $. You will get the answer.

Complete step-by-step answer:
Now taking $\sec \theta =x+\dfrac{1}{4x}$,

We have been given $\sec \theta $ and from that we will find $\tan \theta $.

We know that ${{\sec }^{2}}\theta -1={{\tan }^{2}}\theta $.

So substituting value of $\sec \theta $ in above identity we get,
\[{{\left( x+\dfrac{1}{4x} \right)}^{2}}-1={{\tan }^{2}}\theta \]
Simplifying we get,
\[\begin{align}
 & {{x}^{2}}+2(x)\dfrac{1}{4x}+{{\left( \dfrac{1}{4x} \right)}^{2}}-1={{\tan }^{2}}\theta \\
 & {{x}^{2}}+\dfrac{1}{2}+\left( \dfrac{1}{16{{x}^{2}}} \right)-1={{\tan }^{2}}\theta \\
 & {{x}^{2}}+\left( \dfrac{1}{16{{x}^{2}}} \right)-\dfrac{1}{2}={{\tan }^{2}}\theta \\
 & {{\left( x-\dfrac{1}{4x} \right)}^{2}}={{\tan }^{2}}\theta \\
\end{align}\]

So taking square root of both sides we get,
\[\tan \theta =\pm \left( x-\dfrac{1}{4x} \right)\]
We get,
\[\tan \theta =\left( x-\dfrac{1}{4x} \right)\] and \[\tan \theta =-x+\dfrac{1}{4x}\]

So now we have got $\tan \theta $.
Now adding $\tan \theta $ and $\sec \theta $,
$\sec \theta +\tan \theta =x+\dfrac{1}{4x}\pm \left( x-\dfrac{1}{4x} \right)$

$\sec \theta +\tan \theta =x+\dfrac{1}{4x}+\left( x-\dfrac{1}{4x} \right)$ or $\sec \theta +\tan \theta =x+\dfrac{1}{4x}-\left( x-\dfrac{1}{4x} \right)$

Simplifying we get,
$\sec \theta +\tan \theta =2x$ or $\sec \theta +\tan \theta =\dfrac{1}{2x}$
So we get the values $\sec \theta +\tan \theta =2x$ or $\sec \theta +\tan \theta =\dfrac{1}{2x}$.

Hence proved.

Note: Read the question carefully. Do not make any silly mistakes. Also, you must be familiar with the trigonometric identities. Do not confuse yourself while simplifying. Also, take care that no terms are missing and do not jumble with the signs.