Question

$(secA+\tan A)(1-\sin A)=?$

• A. $CosA$
• B. $\sin A$
• C. $secA$
• D. $cotA$

Answer 1

The correct option is A

$CosA$

Use appropriate trigonometric identities and simplify given expression

We know that,

$$\begin{gathered} secA=\frac{1}{\cos A} \\ \tan A=\frac{\sin A}{\cos A} \end{gathered}$$

Substitute the above values in given expression

$$\begin{gathered} (secA+\tan A)(1–\sin A) \\ =\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\times (1-\sin A) \\ =\left(\frac{1+\sin A}{\cos A}\right)(1–\sin A) \\ =\frac{(1–{\sin }^{2}A)}{\cos A}\left[\because \left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ =\frac{{\cos }^{2}A}{\cos A}\left[\because {\sin }^{2}x+{\cos }^{2}x=1\right] \\ =\cos A \end{gathered}$$

Hence, option A is the correct answer.