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If Cosec A = 2 find `1/(tan A) + (sin A)/(1 + cos A)`
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`Cosec A = "hypotenuse"/"opposite side" = 2/1`
Let x be the adjacent side
By applying Pythagoras theorem
`AC^2 = AB^2 + BC^2`
4 = 1 + ЁЭСе2
`x^2 = 3 => x = sqrt3`
`sin A = 1/(cosec A) = 1/2`
`tan A = (AB)/(BC) = 1/sqrt3`
`cos A = (BC)/(AC) = sqrt3/2`
Substitute in equation we get
`1/tan A + sin A /(1+ cos A) = 1/(1/sqrt3) + (1/2)/(1 + sqrt3/2)`
`=> sqrt3 + (1/2)/((2 + sqrt3)/2) = sqrt3 + 1/(2 + sqrt3) = (2sqrt3 + 3 +1)/(2 + sqrt3) = (2sqrt3 + 4)/(2 + sqrt3) = (2(2 + sqrt3))/(2 + sqrt3) = 2`
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