3. Completo : \( \left(1-\sin ^{2} \theta\right)=(1-\sin \theta)(\cdots) \), hence tho iventits \( \frac{1-\sin ^{2} A}{\cos A\left(\frac{1}{\cos A}+\tan A\right)}=1-\sin A \)

1. Recognize the Pythagorean identity: \[ 1-\sin^{2}\theta = (1-\sin \theta)(1+\sin \theta). \] Thus, the missing factor is \(1+\sin \theta\). 2. Now, consider the expression: \[ \frac{1-\sin ^{2} A}{\cos A\left(\frac{1}{\cos A}+\tan A\right)}. \] Replace the numerator by its factorized form: \[ \frac{(1-\sin A)(1+\sin A)}{\cos A\left(\frac{1}{\cos A}+\tan A\right)}. \] 3. Rewrite \(\tan A\) as \(\frac{\sin A}{\cos A}\). Then, the denominator becomes: \[ \cos A\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right) = \cos A \left(\frac{1+\sin A}{\cos A}\right) = 1+\sin A. \] 4. Substitute the simplified denominator back into the expression: \[ \frac{(1-\sin A)(1+\sin A)}{1+\sin A}. \] 5. Cancel the common factor \((1+\sin A)\): \[ 1-\sin A. \] Thus, we have shown that \[ \frac{1-\sin ^{2} A}{\cos A\left(\frac{1}{\cos A}+\tan A\right)} = 1-\sin A. \]