Question

If $\sec \theta = x + \dfrac{1}{{4x}}$ , then the value of $\sec \theta + \tan \theta$ is equal to

Answer 1

Hint: We can use some of the basic trigonometric formulas which are related to the functions mentioned in the question for example ${\sec ^2}\theta - {\tan ^2}\theta = 1$ use this formula to find the value of an unknown trigonometric function i.e. $\tan\theta$ in terms of ‘x’. THen we just need to add both.

Complete step-by-step answer:
According to the given information $\sec \theta = x + \dfrac{1}{{4x}}$ ---(equation 1)
To find the value of $\sec \theta + \tan \theta$ we need the value of $\tan \theta$
Let $\sec \theta + \tan \theta$ be the equation 2
So by the trigonometric formula ${\sec ^2}\theta - {\tan ^2}\theta = 1$
$\Rightarrow$ ${\tan ^2}\theta = {\sec ^2}\theta - 1$
Now let put the value of $\sec \theta$ by the equation 1
${\tan ^2}\theta = {(x + \dfrac{1}{{4x}})^2} - 1$
$\Rightarrow$ ${\tan ^2}\theta = {(\dfrac{{4{x^2} + 1}}{{4x}})^2} - 1$
$\Rightarrow$ ${\tan ^2}\theta = \dfrac{{16{x^4} + 1 + 8{x^2}}}{{16{x^2}}} - 1$
$\Rightarrow$ ${\tan ^2}\theta = \dfrac{{16{x^4} + 1 + 8{x^2} - 16{x^2}}}{{16{x^2}}}$ $= \dfrac{{16{x^4} + 1 + - 8{x^2}}}{{16{x^2}}}$
$\Rightarrow$ ${\tan ^2}\theta = \dfrac{{{{(4{x^2} - 1)}^2}}}{{16{x^2}}}$
Applying square root on both sides
$\Rightarrow$ $\sqrt {{{\tan }^2}\theta } = \sqrt {\dfrac{{{{(4{x^2} - 1)}^2}}}{{16{x^2}}}}$
$\Rightarrow$ $\tan \theta = \dfrac{{4{x^2} - 1}}{{4x}}$ = $x - \dfrac{1}{{4x}}$
Now put the value of $\sec \theta$ and $\tan \theta$ in equation 2
$\Rightarrow$ $\sec \theta + \tan \theta$ = $x + \dfrac{1}{{4x}}$ + $x - \dfrac{1}{{4x}}$
$\Rightarrow$ $\sec \theta + \tan \theta = 2x$

Note: In these types of questions use the basic trigonometric formula like ${\sec ^2}\theta - {\tan ^2}\theta = 1$ to get the value of $\tan \theta$ then to simplify the value of $\tan \theta$ the simplest form to use the value in the question after finding the value of $\tan \theta$ directly use it and find the value of $\sec \theta + \tan \theta$ .