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`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))` |
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Answer» Correct Answer - 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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