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Question #b29dc

1 Answer

Sep 8, 2016

$cot(1395^@)=-1$ .

Explanation:

$cot(1395^@)=cot(1440^@-45^@)=cot{4(360^@)-45^@}$ .

Therefore, $1395^@$ lies in the $IV^(th)$ Quadrant, in which, $cos$

& $sec$ are $+ve$ , so, cot is $-ve$ , and,

$cot(1395^@)=cot{4(360^@)-45^@}=-cot45^@=-1$ .

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