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The value of 70^0+4cos70^0 is 1/(sqrt(3)...

The value of `70^0+4cos70^0` is `1/(sqrt(3))` (b) `sqrt(3)` (c) `2sqrt(3)` (d) `1/2`

A

`(1)/(sqrt(3))`

B

`sqrt(3)`

C

`2sqrt(3)`

D

`(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`cot70^(@)+4cos70^(@)=(cos70^(@)+4sin70^(@)cos70^(@))/(sin70^(@))`
`=(cos 70^(@)+2sin140^(@))/(sin70^(@))`
`=(cos 70^(@)+2sin(180^(@)-40^(@)))/(sin70^(@))`
`=(sin20^(@)+sin40^(@)+sin40^(@))/(sin70^(@))`
`=(2sin 30^(@)cos(10^(@)+sin40^(@)))/(sin70^(@))`
`=(sin 80^(@)+sin40^(@))/(sin70^(@))`
`=(2sin 60^(@)cos 20^(@))/(sin70^(@))`
`=sqrt(3)`.
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