MCQ
$\sec^210^\circ-\cot^280^\circ=?$
- ✓$1$
- B$0$
- C$\frac{3}{2}$
- D$\frac{1}2{}$
✓
Answer
Correct option: A.
$1$
$\sec^210^\circ-\cot^280^\circ$
$=\sec^210^\circ-\cot^2(90^\circ-10^\circ)$
$=\sec^210^\circ-\tan^210^\circ$
$=1$
$=\sec^210^\circ-\cot^2(90^\circ-10^\circ)$
$=\sec^210^\circ-\tan^210^\circ$
$=1$
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