Question

Which of the following options are equal to $\frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}$ ?

• A. $2cosecA$
• B. $\frac{1}{\sin A}$
• C. $\frac{2}{\sin A}$
• D. $2$

Answer 1

Answer: (A), (C)

The correct options are
A

$2cosecA$

C

$\frac{2}{\sin A}$

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

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

$=\tan A\left(\frac{\sec A+1+\sec A-1}{{\sec }^{2}A-1}\right)$

$=tanA\left(\frac{2\sec A}{{\tan }^{2}A}\right)...(1+{\tan }^{2}A={\sec }^{2}A)$

$=\frac{2\sec A}{\tan A}$

$=\frac{2}{\cos A\left(\frac{\sin A}{\cos A}\right)}....(\tan A=\frac{\sin A}{\cos A})$

$=\frac{2}{\sin A}=2cosecA$ $...(\frac{1}{\sin A}=cosecA)$