MCQ
$(\sec A +\tan A )(1-\sin A )= ?$
- A$\sin A$
- ✓$\cos A$
- C$\sec A$
- D$cosec\ A$
✓
Answer
Correct option: B.
$\cos A$
$(\sec A+\tan A)(1-\sin A)$
$=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)$
$=\left(\frac{1+\sin A}{\cos A}\right)(1-\sin A)$
$=\frac{1-\sin ^2 A}{\cos A}=\frac{\cos ^2 A}{\cos A} \quad\left(\because 1-\sin ^2 \theta=\cos ^2 \theta\right)$
$=\cos A$
$=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)$
$=\left(\frac{1+\sin A}{\cos A}\right)(1-\sin A)$
$=\frac{1-\sin ^2 A}{\cos A}=\frac{\cos ^2 A}{\cos A} \quad\left(\because 1-\sin ^2 \theta=\cos ^2 \theta\right)$
$=\cos A$
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