Question

If $\tan x = x - \dfrac{1}{{4x}}$ then $\sec x - \tan x$ is equal to:

A) 2x

B) -2x

C) 4x

D) -4x

Answer 1

Hint : In this question value of $\tan x$ is given and we know the trigonometric formula ${\tan ^2}x + 1 = {\sec ^2}x$ so by substituting the value of $\tan x$ in the equation we will find the value of function $\sec x$ and then we will find the values of function $\sec x - \tan x$.

Complete step-by-step answer :

Given, $\tan x = x - \dfrac{1}{{4x}} - - (i)$

We know one of the trigonometric Pythagoras theorems

${\tan ^2}x + 1 = {\sec ^2}x$

This can also be written as

${\sec ^2}x = 1 + {\tan ^2}x - - (ii)$

Now we will substitute the values of $\tan x$ from (i) in equation (ii), hence we can write

${\sec ^2}x = 1 + {\left( {x - \dfrac{1}{{4x}}} \right)^2}$

By further solving this obtained equation we get

${\sec ^2}x = 1 + {x^2} + \dfrac{1}{{16{x^2}}} - \dfrac{1}{2}$

${\sec ^2}x = {x^2} + \dfrac{1}{{16{x^2}}} + \dfrac{1}{2} - - (iii)$

Now since we know the square rule ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ , hence we can write the RHS of the equation (iii) as

${\sec ^2}x = {\left( {x + \dfrac{1}{{4x}}} \right)^2}$

Eliminate the square power from both sides, we get

$\sec x = \pm \left( {x + \dfrac{1}{{4x}}} \right) - - (iv)$

Now since we need to find the value of trigonometric function $\sec x - \tan x$ , we will substitute the value of $\sec x = + \left( -{x + \dfrac{1}{{4x}}} \right)$ and $\tan x\left( { = x - \dfrac{1}{{4x}}} \right)$ from equation (i), we get

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

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

$\Rightarrow \sec x - \tan x = \dfrac{2}{{4x}}$

$\Rightarrow \sec x - \tan x = \dfrac{1}{{2x}}$

Now since the obtained value does not match with and of the above options, so now we will substitute the value $\sec x = - \left( {x + \dfrac{1}{{4x}}} \right)$ from equation (iv), we can write

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

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

$\Rightarrow \sec x - \tan x = -x - \dfrac{1}{{4x}} - x +\dfrac{1}{{4x}}$

$\Rightarrow \sec x - \tan x = -2x$

Hence we get the value of the trigonometric function $\sec x - \tan x = -2x$ and this value matches with option (B).

So, the correct answer is “Option A”.

Note : It is interesting to note that in these types of the questions, we should always opt to simplify the expression using trigonometric identities. So, it is very essential for the students to remember each trigonometric identity along with the sign convention.