MCQ
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})=$
- A$\sec\text{A}$
- B$\sin\text{A}$
- C$\text{cosec A}$
- ✓$\cos\text{A}$
✓
Answer
Correct option: D.
$\cos\text{A}$
The given expression is $(\sec\text{A}+\tan\text{A})(1-\sin\text{A})$
Simplifying the given expression, we have
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})$
$=\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A})$
$=\Big(\frac{1+\sin\text{A}}{\cos\text{A}}\Big)\times(1-\sin\text{A})$
$=\frac{(1+\sin\text{A})(1-\sin\text{A})}{\cos\text{A}}$
$=\frac{1-\sin^2\text{A}}{\cos\text{A}}$
$=\frac{\cos^2\text{A}}{\cos\text{A}}$
$=\cos\text{A}$
Therefore, the correct option is $(d).$
Simplifying the given expression, we have
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})$
$=\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A})$
$=\Big(\frac{1+\sin\text{A}}{\cos\text{A}}\Big)\times(1-\sin\text{A})$
$=\frac{(1+\sin\text{A})(1-\sin\text{A})}{\cos\text{A}}$
$=\frac{1-\sin^2\text{A}}{\cos\text{A}}$
$=\frac{\cos^2\text{A}}{\cos\text{A}}$
$=\cos\text{A}$
Therefore, the correct option is $(d).$
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