Question

The expression $\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$ reduces to :

• A. $\frac{1+\sin A}{\cos A}$
• B. $\frac{1-\sin A}{\cos A}$
• C. $\frac{1+\cos A}{\sin A}$
• D. $\frac{1+\cos A}{\cos A}$

Answer 1

The correct option is A $\frac{1+\sin A}{\cos A}$

${\sec }^2\theta =1+{\tan }^2\theta$

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

$\Rightarrow \frac{\tan A+\sec A-({\sec }^{2}A-{\tan }^{2}A)}{\tan A-\sec A+1}$

$\Rightarrow \frac{(\tan A+\sec A)(1-\sec A+\tan A)}{\tan A-\sec A+1}$

$\Rightarrow \tan A+\sec A$

$\Rightarrow \frac{\sin A}{\cos A}+\frac{1}{\cos A}$

$\Rightarrow \frac{1+\sin A}{\cos A}$