# Prove that $$2cos2A+1\over {2cos2A-1}$$ = tan($$60^{\circ}$$ + A)tan($$60^{\circ}$$ – A)

## Solution :

R.H.S. = tan($$60^{\circ}$$ + A)tan($$60^{\circ}$$ – A)

= ($$tan60^{\circ}+tanA\over {1-tan60^{\circ}tanA}$$)($$tan60^{\circ}-tanA\over {1+tan60^{\circ}tanA}$$)

= ($$\sqrt{3}+tanA\over {1-\sqrt{3}tanA}$$)($$\sqrt{3}-tanA\over {1+\sqrt{3}tanA}$$)

= $$3-tan^2A\over{1-3tan^2A}$$ = $$3cos^2A-sin^2A\over {cos^2A-3sin^2A}$$

= $$2cos^2A+cos^2A-2sin^2A+sin^2A\over {2cos^2A-2sin^2A-sin^2A-cos^2A}$$

= $$2(cos^2A-sin^2A)+cos^2A+sin^2A\over {2(cos^2A-sin^2A)-(sin^2A+cos^2A)}$$

= $$2cos2A+1\over {2cos2A-1}$$ = L.H.S

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