The value of [(sec A + tan A) (1−sin A)] is equal to

Here we have to evaluate the value of $(\sec A+\tan A)(1-\sin A)$

Now we are going to solve this

$(\sec A+\tan A)(1-\sin A)$

$=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)$ (here we are using the trigonometrical relation)

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

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

$=\frac{1}{\cos A}\left(1-\sin ^{2} A\right)$ we know that $(a-b)(a+b)=\left(a^{2}-b^{2}\right)$

$=\frac{\cos ^{2} A}{\cos A}$ we use the relation $\sin ^{2} A+\cos ^{2} A=1$

$=\cos A$

Hence the option (c) is correct.