2 2 ~A+1 2 2 ~A-1 = (60^ +A ) (60^ -A )

Question

$\frac{2 \cos 2 \mathrm{~A}+1}{2 \cos 2 \mathrm{~A}-1}=\tan \left(60^{\circ}+\mathrm{A}\right) \tan \left(60^{\circ}-\mathrm{A}\right) $

✓

Answer

\begin{aligned}
& \text { R.H.S. }=\tan \left(60^{\circ}+\mathrm{A}\right) \tan \left(60^{\circ}-\mathrm{A}\right) \\
& =\frac{\sin \left(60^{\circ}+\mathrm{A}\right) \sin \left(60^{\circ}-\mathrm{A}\right)}{\cos \left(60^{\circ}+\mathrm{A}\right) \cos \left(60^{\circ}-\mathrm{A}\right)} \\
& =\frac{2 \sin \left(60^{\circ}+\mathrm{A}\right) \sin \left(60^{\circ}-\mathrm{A}\right)}{2 \cos \left(60^{\circ}+\mathrm{A}\right) \cos \left(60^{\circ}-\mathrm{A}\right)} \\
& =\frac{\cos \left[60^{\circ}+\mathrm{A}-\left(60^{\circ}-\mathrm{A}\right)\right]-\cos \left(60^{\circ}+\mathrm{A}+60^{\circ}-\mathrm{A}\right)}{\cos \left(60^{\circ}+\mathrm{A}+60^{\circ}-\mathrm{A}\right)+\cos \left[60^{\circ}+\mathrm{A}-\left(60^{\circ}-\mathrm{A}\right)\right]} \\
& =\frac{\cos 2 \mathrm{~A}-\cos 120^{\circ}}{\cos 120^{\circ}-\cos 2 \mathrm{~A}} \\
& =\frac{\cos 2 \mathrm{~A}-\cos \left(180^{\circ}-60^{\circ}\right)}{\cos \left(180^{\circ}-60^{\circ}\right)+\cos 2 \mathrm{~A}} \\
& =\frac{\cos 2 \mathrm{~A}-\left(-\cos 60^{\circ}\right)}{-\cos 60^{\circ}+\cos 2 \mathrm{~A}} \\
& =\frac{\cos 2 \mathrm{~A}+\frac{1}{2}}{-\frac{1}{2}+\cos 2 \mathrm{~A}} \\
& =\frac{2 \cos 2 \mathrm{~A}+1}{2 \cos 2 \mathrm{~A}-1} \\
& =\mathrm{L} \cdot \mathrm{H} \cdot \mathrm{S} .
\end{aligned}