$\text{sec}A+\text{tan}A = \frac{5}{2}$ then, $\text{sec}A-\text{tan}A=?$

Question

$\text{sec}A+\text{tan}A = \frac{5}{2}$ then, $\text{sec}A-\text{tan}A=?$

This is my attempt,

$$\text{sec}A +\text{tan}A = \frac{5}{2}$$

$$\implies \text{sec}A = \frac{5}{2} - \text{tan}A$$

$$\implies \text{sec}^2A = \frac{25}{4} -\frac{25}{2}\text{tan}A +\text{tan}^2A$$

$$\implies 1+\text{tan}^2A=\frac{25}{4} -\frac{25}{2}\text{tan}A +\text{tan}^2A$$

$$\implies 1-\frac{25}{4}=-\frac{25}{2}\text{tan}A$$

$$\implies -\frac{21}{4}=-\frac{25}{2}\text{tan}A$$

$$\implies \text{tan}A=\frac{21}{4} \cdot \frac{2}{25}$$

$$\implies \text{tan}A=\frac{21}{50}$$

Now, putting the value,

$$\text{sec}A +\text{tan}A =\frac{5}{2}$$

$$\text{sec}A +\frac{21}{50}=\frac{5}{2}$$

$$\implies \text{sec}A=\frac{52}{25}$$

Now finally,

$$\text{sec}A -\text{tan}A=\frac{52}{25}-\frac{21}{50}$$

$$\implies \text{sec}A -\text{tan}A=\frac{83}{50}$$

But the answer given in my book is $\frac{2}{5}$, so which part of my attempt is wrong?

If there's an problem in my question please inform me. Thanks!

Answer 1

Your third line is incorrect.

Hint: \begin{align} (a - b)^{2} = a^{2} - \mathbf{2ab} + b^{2} \end{align}

Answer 2

Note that $\sin^2x+\cos^2x=1$ gives $\tan^2x+1=\sec^2x$, or $$\sec^2x-\tan^2x=1$$

Now expand it to get $$(\sec x+\tan x)(\sec x-\tan x)=1$$ or $(\frac52)(\sec x -\tan x)=1$ which gives$$\sec x -\tan x = \frac25$$

Answer 3

This is a lengthy process. You can use the identity $\sec^2x - \tan^2x = 1$

Answer 4

There is a shortcut. Since the product $$ \sec^2 A -\tan^2 A = 1,$$ it makes one the reciprocal of the other.

Accordingly reciprocal of $\dfrac52 $ is $\dfrac25. $