Question

$\frac{\left(1-\sin A\right)}{\cos A}$ is equal to

• A. $\frac{\cos A}{\left(1+\sin A\right)}$
• B. $\frac{\sin A}{\left(1-\cos A\right)}$
• C. $\frac{\tan A}{\left(1+\tan A\right)}$
• D. $\frac{\tan A}{\left(1+\cos A\right)}$

Answer 1

The correct option is A

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

Explanation for the correct option.

We have,

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

Multiply numerator and denominator by $\left(1+\sin A\right)$

$$\begin{gathered} \frac{\left(1-\sin A\right)}{\cos A}=\frac{\left(1-\sin A\right)\times \left(1+\sin A\right)}{\cos A\times \left(1+\sin A\right)} \\ =\frac{\left(1-{\sin }^{2}A\right)}{\cos A\left(1+\sin A\right)}...\left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{{\cos }^{2}A}{\cos A\left(1+\sin A\right)}...\left[\because 1-si{n}^{2}A=co{s}^{2}A\right] \\ =\frac{\cos A}{\left(1+\sin A\right)} \end{gathered}$$

Therefore, $\frac{\left(1-\sin A\right)}{\cos A}=\frac{\cos A}{\left(1+\sin A\right)}$.

Hence, the correct option is $A$.