MCQ
$3\cos^260^\circ+2\cot^230^\circ-5\sin^245^\circ=?$
- A$\frac{13}{6}$
- ✓$\frac{17}{4}$
- C$1$
- D$4$
✓
Answer
Correct option: B.
$\frac{17}{4}$
$3\cos^260^\circ+2\cot^230^\circ-5\sin^245^\circ$
$=3\times\Big(\frac{1}{2}\Big)^2+2\big(\sqrt3\big)^2-5\Big(\frac{1}{\sqrt2}\Big)^2$
$=3\times\frac{1}{4}+2\times3-5\times\frac{1}{2}$
$=\frac{3}{4}+6-\frac{5}{2}$
$=\frac{3+24-10}{4}$
$=\frac{17}{4}$
$=3\times\Big(\frac{1}{2}\Big)^2+2\big(\sqrt3\big)^2-5\Big(\frac{1}{\sqrt2}\Big)^2$
$=3\times\frac{1}{4}+2\times3-5\times\frac{1}{2}$
$=\frac{3}{4}+6-\frac{5}{2}$
$=\frac{3+24-10}{4}$
$=\frac{17}{4}$
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