MCQ
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})=?$
- A$\sin\text{A}$
- ✓$\cos\text{A}$
- C$\sec\text{A}$
- D$\text{cosec}\text{A}$
✓
Answer
Correct option: B.
$\cos\text{A}$
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})$
$=\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A})$
$=\frac{(1+\sin\text{A})}{\cos\text{A}}\times(1-\sin\text{A})$
$=\frac{1-\sin^2\text{A}}{\cos\text{A}}$
$=\frac{\cos^2\text{A}}{\cos\text{A}}$
$=\cos\text{A}$
$=\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A})$
$=\frac{(1+\sin\text{A})}{\cos\text{A}}\times(1-\sin\text{A})$
$=\frac{1-\sin^2\text{A}}{\cos\text{A}}$
$=\frac{\cos^2\text{A}}{\cos\text{A}}$
$=\cos\text{A}$
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