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I was trying to simplify $$\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1}$$

I change all $\tan, \cot, \sec, \operatorname{cosec}$ into $\cos$ and $\sin$ using conversion formulas (I find it easier). Here, I get this:

$$\frac{\sin A + 1 - \cos A}{\sin A - 1 + \cos A}$$ from which I can not simplify into anything. I have tried "rationalization technique" by multiplying both numerator and denominator by $\sin A \pm (1 - \cos A)$ but to no yield.

I was then surprised to know the model solution, which is very clever, we replace the $1$ in the numerator with $$\sec^2 A - \tan^2 A$$ to get $$\frac{(\tan A + \sec A) - (\sec^2 A -\tan^2 A)}{\tan A - \sec A + 1}$$ after which it is just one step factoring and cancelling and answer is $$\frac{1 + \sin A}{\cos A}$$

For a person who has only solved these problems by first converting into $\sin, \cos$, I find it difficult to comprehend. Moreover, I also really find it difficult how to think this solution in an exam. I know that that I can not think of such clever solution.

So, I was wondering if there is any easier solution. Any help would be really appreciated.

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    $\begingroup$ $\tan^2 x + 1 = \sec^2 x$ is a basic trigonometric identity, which is equivalent to $sin^2 x + \cos ^2 x = 1$, and divide the entire equation by $\cos^2 x$ $\endgroup$
    – KingLogic
    Commented Aug 19, 2021 at 18:15
  • $\begingroup$ Hi @KingLogic, I am aware of this identity, however, I do not like working with $\tan$ or $\sec$, I like to work with $\cos$ and $\sin$ only, by converting everything. Whether it's just my matter of taste or a bad practice, I don't know? $\endgroup$
    – MangoPizza
    Commented Aug 20, 2021 at 3:26
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    $\begingroup$ Note that the original expression $\displaystyle\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1}$ is undefined at $270^\circ$ and $0^\circ,$ whereas your initial simplification $\displaystyle\frac{\sin A + 1 - \cos A}{\sin A - 1 + \cos A}$ equals $0$ at $270^\circ$ and your subsequent simplification $\displaystyle\frac{1 + \sin A}{\cos A}$ equals $1$ at $0^\circ.$ My explanation is here. $\endgroup$
    – ryang
    Commented Dec 8, 2021 at 15:09

2 Answers 2

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Based on how your proceeded, you could have multiplied numerator and denominator by $\sin A + \cos A + 1$,

Then, $\displaystyle \frac{1 + \sin A - \cos A}{-1 + \sin A + \cos A} = \frac{((1 + \sin A) - \cos A) ((1 + \sin A) + \cos A)}{((\sin A + \cos A) - 1) ((\sin A + \cos A) + 1)} $

$ = \cfrac{1 + \sin^2 A + 2 \sin A - \cos^2A}{2 \sin A \cos A}$

$ = \cfrac{2 \sin A (1 + \sin A)}{2 \sin A \cos A} = \cfrac{1 + \sin A}{\cos A}$

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Another way. $$LS=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}+2\sin^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}-2\sin^2\frac{A}{2}}=\frac{\cos\frac{A}{2}+\sin\frac{A}{2}}{\cos\frac{A}{2}-\sin\frac{A}{2}}=$$ $$=\frac{\left(\cos\frac{A}{2}+\sin\frac{A}{2}\right)^2}{\cos^2\frac{A}{2}-\sin^2\frac{A}{2}}=\frac{1+\sin A}{\cos A}.$$

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