Question

What is $\frac{1+\sin A}{1-\sin A}-\frac{1-\sin A}{1+\sin A}$ equal to ?

• A. $\sec A\tan A$
• B. $2\sec A.\tan A$
• C. $4\sec A.\tan A$
• D. $4\text{cosec}A.\cot A$

Answer 1

The correct option is C $4\sec A.\tan A$

$\frac{(1+\sin A)}{(1-\sin A)}-\frac{(1-\sin A)}{(1+\sin A)}=\frac{(1+\sin A)(1+\sin A)-(1-\sin A)(1-\sin A)}{(1-\sin A)(1+\sin A)}=\frac{(1+2\sin A+{\sin }^{2}A)-(1-2\sin A+{\sin }^{2}A)}{(1-{\sin }^{2}A)}$

$=\frac{4\sin A}{{\cos }^{2}A}$ ......... $[\because 1-{\sin }^{2}A={\cos }^{2}A]$

$=4\tan A\sec A$

Hence, C is correct.