Step 1: Write in terms of $\sin A$ and $\cos A$
\[
\sec A = \frac{1}{\cos A}, \tan A = \frac{\sin A}{\cos A}
\]
So,
\[
\sec A + \tan A = \frac{1 + \sin A}{\cos A}
\]
Step 2: Multiply with $(1 - \sin A)$
\[
(\sec A + \tan A)(1 - \sin A) = \frac{1 + \sin A}{\cos A} \times (1 - \sin A)
\]
\[
= \frac{(1 + \sin A)(1 - \sin A)}{\cos A}
\]
\[
= \frac{1 - \sin^2 A}{\cos A}
\]
Step 3: Simplify using identity
Since $1 - \sin^2 A = \cos^2 A$:
\[
= \frac{\cos^2 A}{\cos A} = \cos A
\]
Step 4: Conclusion
Thus,
\[
(\sec A + \tan A)(1 - \sin A) = \cos A
\]
The correct answer is option (D).
The value of $\dfrac{1+\cot^2 \theta}{1+\tan^2 \theta}$ will be:
If $3 \cot A = 4$, then the value of $\dfrac{1 - \tan^2 A}{1 + \tan^2 A}$ will be: