Question

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

Answer 1

$$\tan (60 + A) \cdot \tan (60 - A) = 2\cos 2A + \frac{1}{2\cos 2A - 1}$$ $$\tan (60 + A) \tan (60 - A) = 2\cos 2A + \frac{1}{2\cos 2A - 1}$$

We have

$$\tan (60 + A) = \frac{\tan 60 + \tan A}{1 - \tan 60 \tan A} = \frac{\sqrt{3} + \tan A}{1 - \sqrt{3} \tan A}$$ $$\tan (60 - A) = \frac{\tan 60 - \tan A}{1 + \tan 60 \tan A} = \frac{\sqrt{3} - \tan A}{1 + \sqrt{3} \tan A}$$ $$\tan (60 + A) \tan (60 - A) = \frac{\sqrt{3} + \tan A}{1 - \sqrt{3} \tan A} \cdot \frac{\sqrt{3} - \tan A}{1 + \sqrt{3} \tan A} = \frac{3 - \tan^2 A}{1 - 3 \tan^2 A} = \frac{3 - \frac{\sin^2 A}{\cos^2 A}}{1 - 3 \frac{\sin^2 A}{\cos^2 A}} = \frac{3 - \frac{\sin^2 A}{\cos^2 A}}{1 - 3 \frac{\sin^2 A}{\cos^2 A}} = \frac{3 \cos^2 A - \sin^2 A}{\cos^2 A - 3 \sin^2 A} = \frac{2 \cos^2 A + \cos 2A}{\cos 2A - 2 \sin^2 A} = \frac{1 + \cos 2A + \cos 2A}{\cos 2A - 1 + \cos 2A} = \frac{1 + 2 \cos 2A}{-1 + 2 \cos 2A} = \frac{1}{2 \cos 2A - 1} + \frac{2 \cos 2A}{2 \cos 2A - 1}$$

Thus we have

$$\tan (60 + A) \tan (60 - A) = \frac{1}{2\cos 2A - 1} + \frac{2\cos 2A}{2\cos 2A - 1}.$$

And identity $\tan (60 + A) \tan (60 - A) = 2\cos 2A + \frac{1}{2\cos 2A - 1}$ is wrong, because for $A = 60$

$$\tan (60 + 60) \tan (60 - 60) = \tan 120 \tan 0 = 0 \neq -\frac{3}{2} = 2\cos 120 + \frac{1}{2\cos 120 - 1} = -1 + \frac{1}{2}$$