Question

the value of cos(60+2A)÷cos 2A-√3 sin 2A

Answer 1

Answer:

Step-by-step explanation:

The simplest way to proceed with such kind of problems is to go with the method of substitution

Here I am substituting A=0

cos0=1

sin0=0

cos60=1÷2

cos(60+2A)÷cos 2A-√3 sin 2A= 1÷2

Answer 2

value of cos(60+2A)÷cos 2A-√3 sin 2A = 1/2

Step-by-step explanation:

cos(60+2A)÷ (cos 2A-√3 sin 2A)

Using Cos ( x + y) = CosxCosy - SinxSiny

cos(60+2A) = Cos60 Cos2A - Sin60Sin2A

Cos60 = 1/2

Sin60 = √3/2

=>  cos(60+2A) =  (1/2) Cos2A - (√3/2)Sin2A

=> cos(60+2A) = (cos 2A-√3 sin 2A) / 2

using this value

cos(60+2A)÷ (cos 2A-√3 sin 2A)

= ( (cos 2A-√3 sin 2A) / 2) ÷ (cos 2A-√3 sin 2A)

= 1/2

but cos 2A-√3 sin 2A ≠ 0

=> √3 sin 2A ≠  cos 2A

=> Tan2A ≠ 1/√3

=>  2A ≠ nπ + 30°

=> A ≠ nπ/2 + 15°  

cos(60+2A)÷ (cos 2A-√3 sin 2A) = 1/2