Question

Let $0 < x < \dfrac{\pi }{4}$ , then $\sec 2x - \tan 2x$ is equal to
A) $\tan \left( {x - \dfrac{\pi }{4}} \right)$
B) $\tan \left( {\dfrac{\pi }{4} - x} \right)$
C) $\tan \left( {x + \dfrac{\pi }{4}} \right)$
D) $\tan \left( {\dfrac{\pi }{4} + x} \right)$

Answer 1

Hint : In this question, we have to evaluate the given trigonometric expression. Here, we will use trigonometric and algebraic identities in order to simplify the given expression and convert the given terms in terms of $\tan x$ , which helps us to get the required answer.

Complete step-by-step answer :
The given trigonometric expression is $\sec 2x - \tan 2x$ .
As we know that,
$\cos x = \dfrac{1}{{\sec x}}$
$\Rightarrow \sec x = \dfrac{1}{{\cos x}}$
Therefore, $\sec 2x = \dfrac{1}{{\cos 2x}}$
And, also we know that,
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
Therefore, $\tan 2x = \dfrac{{\sin 2x}}{{\cos 2x}}$
Applying the values of $\sec 2x$ and $\tan 2x$ in $\sec 2x - \tan 2x$ , we get,
$= \dfrac{1}{{\cos 2x}} - \dfrac{{\sin 2x}}{{\cos 2x}}$
= $\dfrac{{1 - \sin 2x}}{{\cos 2x}}$
Now, as we know,
${\sin ^2}x + {\cos ^2}x = 1$
And, $\sin 2x = 2\sin x\cos x$
Also, $\cos 2x = {\cos ^2}x - {\sin ^2}x$
Now, substituting these values, we get,
= $\dfrac{{{{\sin }^2}x + {{\cos }^2}x - 2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}$
Now, we know that,
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
Here, the numerator is in the form of the above property. Therefore, we have,
= $\dfrac{{{{\left( {\cos x - \sin x} \right)}^2}}}{{\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}$
Where, $a = \cos x$ and $b = \sin x$
Here, we can further simplify it by expanding the denominator,
As, we see the denominator is in the property,
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
Therefore, we have,
$= \dfrac{{{{\left( {\cos x - \sin x} \right)}^2}}}{{\left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right)}}$
Where, $a = \cos x$ and $b = \sin x$
As, $\cos x - \sin x$ is common in both numerator and denominator. Now, cancel out the terms.
We have,
= $\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}$
Dividing the numerator and denominator by $\cos x$ , in order to get the values in terms of $\tan x$ ,
$= \dfrac{{\dfrac{{\cos x}}{{\cos x}} - \dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{{\cos x}}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}}}}$
On cancelling and substituting $\tan x = \dfrac{{\sin x}}{{\cos x}}$ , we get,
= $\dfrac{{1 - \tan x}}{{1 + \tan x}}$
We know from the trigonometric table that,
$\tan \dfrac{\pi }{4} = 1$
Therefore,
= $\dfrac{{\tan \dfrac{\pi }{4} - \tan x}}{{1 + \tan \dfrac{\pi }{4}\tan x}}$
This is in the form of the trigonometric identity,
$\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$
Where, $a = \dfrac{\pi }{4}$ and $b = x$
$\Rightarrow \sec 2x - \tan 2x = \tan \left( {\dfrac{\pi }{4} - x} \right)$
So, the correct answer is “Option B”.

Note : In this question, it is important to note that whenever these types of questions are given, be clear and confident about the identities which helps the simplification process. However, at this step $\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}$ , instead of dividing the numerator and denominator by $\cos x$ , we can divide the numerator and the denominator by $\sqrt 2$ . Then, by using the values from trigonometric table and also using the trigonometric identities $sin\left( {a - b} \right) = \sin a\cos b - \cos a\sin b$ and $\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$ we can reach the solution.