Question

Evaluate
$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$

• A. $\sec A+\tan A$
• B. $\sin A$
• C. $1$
• D. $0$

Answer 1

The correct option is A $\sec A+\tan A$

The value of $\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$ is equal to
$=\frac{\tan A+\sec A-({\sec }^{2}A-{\tan }^{2}A)}{\tan A-\sec A+1}$
$=\frac{(\tan A+\sec A)-(\sec A+\tan A)(\sec A-\tan A)}{\tan A-\sec A+1}$
$=\frac{(\tan A+\sec A)(1-\sec A+\tan A)}{\tan A-\sec A+1}$
$=\tan A+\sec A$
$=\frac{\sin A}{\cos A}+\frac{1}{\cos A}=\frac{\sin A+1}{\cos A}=\frac{1+\sin A}{\cos A}$