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.
Best Answer
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}$