^230^ +4 ^245^ - ^260^ =? - Vidyadip

MCQ

$\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ=?$

A
$0$
✓
$\frac{1}{4}$
C
$4$
D
$1$

✓

Answer

Correct option: B.

$\frac{1}{4}$

$\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ$
$=\Big(\frac{1}{2}\Big)^2+4(1)^2-(2)^2$
$=\frac{1}{4}+4-4$
$=\frac{1}4{}$

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