MCQ
$\frac{2\sin^263^\circ+1+2\sin^227^\circ}{3\cos^217^\circ-2+3\cos^273^\circ}=?$
- A$\frac32$
- B$\frac23$
- C$2$
- ✓$3$
✓
Answer
Correct option: D.
$3$
$\frac{2\sin^263^\circ+1+2\sin^227^\circ}{3\cos^217^\circ-2+3\cos^273^\circ}$
$=\frac{2\sin^263^\circ+2\sin^227+1}{3\cos^217^\circ+3\cos^273^\circ-2}$
$=\frac{2\sin^263^\circ+2\cos^263^\circ+1}{3\cos^217^\circ+3\sin^217^\circ-2}$
$=\frac{2\times1+1}{3\times1-2}$
$=\frac{2+1}{3-2}$
$=3$
$=\frac{2\sin^263^\circ+2\sin^227+1}{3\cos^217^\circ+3\cos^273^\circ-2}$
$=\frac{2\sin^263^\circ+2\cos^263^\circ+1}{3\cos^217^\circ+3\sin^217^\circ-2}$
$=\frac{2\times1+1}{3\times1-2}$
$=\frac{2+1}{3-2}$
$=3$
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