Question

Prove that:

$tan\theta tan({60}^{\circ }-\theta )tan({60}^{\circ }+\theta )$

Answer 1

We have,
$LHS=tan\theta tan({60}^{\circ }-\theta )tan({60}^{\circ }+\theta )=\frac{sin\theta ({60}^{\circ }-\theta )sin({60}^{\circ }+\theta )}{cos\theta cos({60}^{\circ }-\theta )cos({60}^{\circ }+\theta )}=\frac{2sin\theta sin({60}^{\circ }-\theta )sin({60}^{\circ }-\theta )}{cos\theta cos({60}^{\circ }-\theta )cos({60}^{\circ }+\theta )}=\frac{sin\theta \left[2sin\right({60}^{\circ }-\theta \left)sin\right({60}^{\circ }+\theta \left)\right]}{cos\theta \left[2cos\right({60}^{\circ }-\theta \left)cos\right({60}^{\circ }+\theta \left)\right]}=\frac{sin\theta [cos\left\{\right({60}^{\circ }-\theta )-({60}^{\circ }+\theta \left)\right\}-cos\left\{\right({60}^{\circ }-\theta )+({60}^{\circ }+\theta \left)\right\}]}{cos\theta \left[cos\left\{\right({60}^{\circ }-\theta )+({60}^{\circ }+\theta \left)\right\}\right]+\left[cos\left\{\right({60}^{\circ }-\theta )-({60}^{\circ }+\theta \left)\right\}\right]}=\frac{sin\theta \left[cos\theta \right(-2{\theta }^{\circ })-cos{120}^{\circ }]}{cos\theta [cos{120}^{\circ }+cos(-2{\theta }^{\circ }\left)\right]}=\frac{sin\theta [cos2\theta -cos{120}^{\circ }]}{cos\theta [cos{120}^{\circ }+cos2\theta ]}[\because cos(-\theta )=cos\theta ]=\frac{sin\theta [cos2\theta +sin{30}^{\circ }]}{cos\left[cos\right({90}^{\circ }+{30}^{\circ })+cos2\theta ]}=\frac{sin\theta [cos2\theta +sin{30}^{\circ }]}{cos\theta [-sin{30}^{\circ }+cos2\theta ]}$ [$\because$ cos is negative in IInd quadrant]
$=\frac{sin\theta [cos2\theta +\frac{1}{2}]}{cos[\frac{-1}{2}+cos2\theta ]}=\frac{sin\theta cos2\theta +\frac{1}{2}sin\theta }{\frac{-1}{2}cos\theta +cos\theta cos2\theta }=\frac{sin\theta cos2\theta +\frac{1}{2}sin\theta }{cos\theta cos2\theta -\frac{1}{2}cos\theta }=\frac{\frac{1}{2}\left[2sin\theta cos2\theta \right]+\frac{1}{2}sin\theta }{\frac{1}{2}\left[2cos\theta cos2\theta \right]-\frac{1}{2}cos\theta }=\frac{\frac{1}{2}[2sin\theta cos2\theta +sin\theta ]}{\frac{1}{2}[2cos\theta cos2\theta -cos\theta ]}=\frac{[sin3\theta +sin(-\theta )+sin\theta ]}{[cos3\theta +cos)-\theta ]-cos\theta }=\frac{sin3\theta -sin\theta +sin\theta }{cos3\theta +cos\theta -cos\theta }[\because sin(-\theta )=-sin\theta andcos(-\theta )=cos\theta ]=\frac{sin3\theta }{cos3\theta }=tan3\theta =RHS\therefore tan\theta tan({60}^{\circ }-\theta )tan({60}^{\circ }+\theta )=tan3\theta$.