Question

prove that tan(60 +A ) tan (60-A) =2 cos2A+1/2cos2A-1​

Answer 1

Step-by-step explanation:

Let us use A instead of theeta.

LHS = tan(60°+A)tan(60°-A)

={(tan60°+tanA)/(1-tan60°tanA)}×

{(tan60°-tanA)/(1+tan60°tanA)}

=(tan^2 60° - tan^2 A)/(1-tan^2 60°×tan^2A)

= (3-tan^2 A)/(1–3tan^2 A)

= (3 - sin^2A/cos^2A)/(1 - 3sin^2A/cos^2A)

=(3cos^2 A -sin^2A)/(cos^2 A - 3sin^2A)

={3cos^2A-(1-cos^2A)}/

{cos^2A-3(1-cos^2A)

= (4cos^2 A -1)/(4cos^2 A -3)

={4(1+cos2A)/2–1}/{4(1+cos2A)/2 -3}

= {2(1+cos2A)-1}{2(1+cos2A)-3}

= (2cos2A+1)/(2cos2A-1)= RHS

Answer 2

Answer:

please tell the value of A to prove