Question : If $0°<A<90°$, the value of $\frac{\tan A\ -\ \sec A\ -\ 1}{\tan A\ +\ \sec A\ +\ 1}$ is:
Option 1: $\frac{\sin A-1}{\cos A}$
Option 2: $\frac{1-\sin A}{\cos A}$
Option 3: $\frac{1-\cos A}{\sin A}$
Option 4: $\frac{\sin A+1}{\cos A}$
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Correct Answer: $\frac{\sin A-1}{\cos A}$
Solution : Given: $\frac{\tan A-\sec A-1}{\tan A+\sec A+1}$ = $\frac{\tan A-\sec A-(\sec^{2} A-\tan^{2} A)}{\tan A+\sec A+1}$ = $\frac{(\tan A-\sec A)-(\sec A-\tan A)(\sec A+\tan A)}{\tan A+\sec A+1}$ = $\frac{(\tan A-\sec A)+(\tan A-\sec A)(\sec A+\tan A)}{\tan A+\sec A+1}$ = $\frac{(\tan A-\sec A)(1+\sec A+\tan A)}{\tan A+\sec A+1}$ = $\tan A-\sec A$ = $\frac{\sin A}{\cos A}-\frac{1}{\cos A}$ = $\frac{\sin A-1}{\cos A}$ Hence, the correct answer is $\frac{\sin A-1}{\cos A}$.
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