Question

1/Cos290 + 1/(sqrt3Sin250)

a" "sqrt3/4

b" "4/sqrt3

c" "2/sqrt3

d" "sqrt3/2

Solution

We can solve this problem using basic trigonometric formulas.

In the given equation, 1/Cos290 + 1/(sqrt3Sin250)....(i)

We can write, Cos290 = Cos(270+20) = Sin20

Similarly, Sin250 = Sin(270-20) = -Cos20

Hence, equation (i) can be written as 1/Cos(270+20) + 1/(sqrt3Sin(270-20)

= 1/Sin20 + 1/(sqrt3(-Cos20))

= 1/Sin20 - 1/(sqrt3Cos20)

= (sqrt3Cos20 - Sin20)/(sqrt3Cos20Sin20)

Now multiply and divide above equation with 2.

We get,

= 2/2((sqrt3Cos20 - Sin20))/(sqrt3Sin20Cos20)

= 2((sqrt3Cos20 - Sin20))/(sqrt3*2Sin20Cos20)

We know that, Sin2A = 2SinA*CosA

Hence,

= 2((sqrt3Cos20 - Sin20))/(sqrt3Sin40)

Now again multiply and divide only numerator with 2, we will get

= 2(2/2(sqrt3Cos20 - Sin20))/(sqrt3Sin40)

= 2*2(((sqrt3)/2Cos20 - 1/2Sin20)/(sqrt3Sin40))

= 2*2((Cos30Cos20 - Sin30Sin20)/(sqrt3Sin40))

Using formula Cos(A+B) = CosACosB - SinASinB at the numerator

We get,

= 2*2(Cos50/(sqrt3Sin40))

Also, Cos50 = Cos(90-40) = Sin40

Hence above equation will be,

= 2*2(Sin40/(sqrt3Sin40))

= (2*2)/sqrt3

= 4/sqrt3