Question

The value of $\left[\right(\sec A+\tan A\left)\right(1-\sin A\left)\right]$ is equal to:

• A. ${\tan }^{2}A$
• B. ${\sin }^{2}A$
• C. $\cos A$
• D. $\sin A$

Answer 1

Answer: (C)

$\cos A$

We need to find value of $(\sec A+\tan A)(1-\sin A)$

Putting the values of $\sec A$ and $\tan A$ as

$\sec A=\frac{1}{\cos A}$

$\tan A=\frac{\sin A}{\cos A}$

$(\frac{1}{\cos A}+\frac{\sin A}{\cos A})(1-\sin A)$

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

$=\frac{1-{\sin }^{2}A}{\cos A}$

$=\frac{{\cos }^{2}A}{\cos A}$ .... (As $1-{\sin }^{2}A={\cos }^{2}A)$

$=\cos A$