# Demonstrate 1/sin 10 degrees - root3/cos 10 degrees=4

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### 1 Answer

You need to test the identity 1/sin 10^o - sqrt3/cos 10^o = 4, hence, you should start bringing the fractions to a common denominator, such that:

(cos10^o - sqrt3sin 10^o)/(sin 10^o*cos 10^o) = 4

You should compare the product sin 10^o*cos 10^o to the double angle formula such that:

sin 2alpha = 2 sin alpha*cos alpha

You need to divide by 2 both sides, such that:

(cos10^o - sqrt3sin 10^o)/(2sin 10^o*cos 10^o) = 4/2

(1/2(cos10^o - sqrt3sin 10^o))/(sin 10^o*cos 10^o) = 2

((1/2)cos10^o - ((sqrt3)/2)sin 10^o)/(sin 10^o*cos 10^o) = 2

You may replace sin 30^o for 1/2 and cos 30^o for sqrt3/2, such that:

(sin 30^o*cos10^o - cos 30^o*sin 10^o)/(sin 10^o*cos 10^o) = 2

Using the formula sin(a - b) = sin a*cos b - sin b*cos a, yields:

sin(30^o - 10^o) = 2 sin 10^o*cos 10^o

sin 20^o = 2 sin 10^o*cos 10^o

Hence, testing if the given identity holds yields the valid formula sin 20^o = 2 sin 10^o*cos 10^o, hence, the identity 1/sin 10^o - sqrt3/cos 10^o = 4 holds.