MCQ
Choose the correct option. Justify your choice : $(\sec\text{A}+\tan\text{A})(1-\sin\text{A})= $
- A$\sec \text{A}$
- B$\sin \text{A}$
- C$\text{cosec A}$
- ✓$\cos \text{A}$
✓
Answer
Correct option: D.
$\cos \text{A}$
$(\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}) $
$=\Big(\frac{1+\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A}) $
$=\frac{1-\sin^2\text{A}}{\cos\text{A}} \ \ [\text{since} (\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2]$
$=\frac{\cos^2\text{A}}{\cos\text{A}} $
$=\cos\text{A} \ \ [\because1-\sin^2 \text{A}=\cos^2 \text{A}]$
$=\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A}) $
$=\Big(\frac{1+\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A}) $
$=\frac{1-\sin^2\text{A}}{\cos\text{A}} \ \ [\text{since} (\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2]$
$=\frac{\cos^2\text{A}}{\cos\text{A}} $
$=\cos\text{A} \ \ [\because1-\sin^2 \text{A}=\cos^2 \text{A}]$
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