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 $ .