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cos e c(360^0)/7+cos e c(540^0)/7= cos ...

`cos e c(360^0)/7+cos e c(540^0)/7=` `cos e c(180^0)/7` (b) `cos e c(90^0)/7` `sec(180^0)/7` (d) `sec(90^0)/7`

A

`co sec((180^(@))/(7))`

B

`co sec((90^(@))/(7))`

C

`sec( (180^(@))/(7))`

D

`sec((90^(@))/(7))`

Text Solution

Verified by Experts

The correct Answer is:
A

Let `alpha=(180^(@))/(7)`
`rArr 180^(@)=7alpha`
`rArr 3alpha=180^(@)-4alpha`
`rARr sin 3alpha=sin4alpha`
`therefore co sec ((360^(@))/(7))+co sec((540^(@))/(7))`
`=co sec 2alpha+co sec 3alpha`
`=(sin 3alpha+2 alpha)/(sin 3alpha sin 2alpha)`
`=(sin 4alpha+sin 2alpha)/(sin 3alpha sin 2alpha)`
`=(2sin 3alpha cos alpha)/(sin 3alpha. sin 2alpha)`
`=(2cos alpha)/(2sin alpha. cos alpha)`
`=co sec alpha`
`=co sec ((180^(@))/(7))`
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