Question

Let $0<x<\frac{\pi }{4}$ then $\sec \left(2x\right)-\tan \left(2x\right)$ equals

• A. $\tan \left(x-\frac{\pi }{4}\right)$
• B. $\tan \left(\frac{\pi }{4}-x\right)$
• C. $\tan \left(x+\frac{\pi }{4}\right)$
• D. ${\tan }^2\left(x+\frac{\pi }{4}\right)$

Answer 1

The correct option is B

$\tan \left(\frac{\pi }{4}-x\right)$

Explanation for the correct option:

Step 1: Evaluating the given integration

Given $\sec \left(2x\right)-\tan \left(2x\right)$

Writing the given equation in terms of $\sin \left(x\right)$ and $\cos \left(x\right)$ function

$$\begin{gathered} \sec \left(2x\right)-\tan \left(2x\right)=\frac{1}{\cos \left(2x\right)}-\frac{\sin \left(2x\right)}{\cos \left(2x\right)} \\ =\frac{1-\sin \left(2x\right)}{\cos \left(2x\right)} \\ =\frac{{\sin }^2\left(x\right)+{\cos }^2\left(x\right)-2\sin \left(x\right)\cos \left(x\right)}{{\cos }^2\left(x\right)-{\sin }^2\left(x\right)}\mathbf{}\left[\because sin\left(2x\right)=2sin\left(x\right)cos\left(x\right);si{n}^2\left(x\right)+co{s}^2\left(x\right)=1;cos\left(2x\right)=co{s}^2\left(x\right)-si{n}^2\left(x\right)\right] \\ =\frac{{\left(\cos \left(x\right)-\sin \left(x\right)\right)}^2}{{\cos }^2\left(x\right)-{\sin }^2\left(x\right)} \\ =\frac{{\left(\cos \left(x\right)-\sin \left(x\right)\right)}^2}{\left(\cos \left(x\right)-\sin \left(x\right)\right)\left(\cos \left(x\right)+\sin \left(x\right)\right)}\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right] \\ =\frac{\cos \left(x\right)-\sin \left(x\right)}{\cos \left(x\right)+\sin \left(x\right)} \end{gathered}$$

Step 2: Dividing numerator and denominator by $\cos \left(x\right)$

$$\begin{gathered} \sec \left(2x\right)-\tan \left(2x\right)=\frac{1-\tan \left(x\right)}{1+\tan \left(x\right)} \\ =\frac{\tan \left({\displaystyle \frac{\pi }{4}}\right)-\tan \left(x\right)}{1+\tan \left({\displaystyle \frac{\pi }{4}}\right)\tan \left(x\right)}\left[\because tan\left(\frac{\pi }{4}\right)=1\right] \\ \sec \left(2x\right)-\tan \left(2x\right)=\tan \left(\frac{\pi }{4}-x\right)\left[\because tan\left(A-B\right)=\frac{tan\left(A\right)-tan\left(B\right)}{1+tan\left(A\right)tan\left(B\right)}\right] \end{gathered}$$

Therefore the given expression is equal to $\tan \left(\frac{\pi }{4}-x\right)$

Hence, option (B) is the correct answer.