Basic Trigonometry

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!!

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