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(cos 80^circ)/(sin 10^circ) + cos 59^circ cosec 31 = ______. - Mathematics

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Question

`(cos 80^circ)/(sin 10^circ) + cos 59^circ  "cosec"  31` = ______.

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Solution

`(cos 80^circ)/(sin 10^circ) + cos 59^circ  "cosec"  31` = 2.

Explanation:

`(cos 80^circ)/(sin 10^circ) + cos 59^circ  "cosec"  31^circ`

= `(cos(90^circ - 10^circ))/(sin 10^circ) + cos(90^circ - 31^circ) xx 1/(sin 31^circ)`

= `(sin 10^circ)/(sin 10^circ) + sin 31^circ xx 1/(sin 31^circ)`

= 1 + 1

= 2

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2019-2020 (March) Standard - Delhi set 1
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