Question

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

Answer 1

I hope you will be able to understand

Answer 2

Answer:

To prove: $tan\,(60+A)\:tan\,(60-A)=\frac{2cos\,2A+1}{2cos\,2A-1}$

Consider,

LHS

$=tan\,(60+A)\:tan\,(60-A)$

$=(\frac{tan\,60+tan\,A}{1-tan\,60\:tan\,A})(\frac{tan\,60-tan\,A}{1+tan\,60\:tan\,A})$

$=(\frac{\sqrt{3}+tan\,A}{1-\sqrt{3}\:tan\,A})(\frac{\sqrt{3}-tan\,A}{1+\sqrt{3}\:tan\,A})$

$=\frac{3-tan^2\,A}{1-3\:tan^2\,A}$

RHS

$=\frac{2cos\,2A+1}{2cos\,2A-1}$

$=\frac{2\frac{1-tan^2\,A}{1+tan^2\,A}+1}{2\frac{1-tan^2\,A}{1+tan^2\,A}-1}$

$=\frac{2-2tan^2\,A+1+tan^1\,A}{1+tan^2\,A}\div\frac{2-2tan^2\,A-1-tan^2\,A}{1+tan^2\,A}$

$=\frac{3-tan^1\,A}{1+tan^2\,A}\times\frac{1+tan^2\,A}{3-tan^2\,A}$

$=\frac{3-tan^1\,A}{3-tan^2\,A}$

LHS = RHS

Hence Proved.