Trigonometry – II — Maths STD 11 Science — Question

\begin{aligned}
& \text { L.H.S. }=\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ} \\
& =\frac{\sin 6^{\circ}}{\cos 6^{\circ}} \cdot \frac{\sin 42^{\circ}}{\cos 42^{\circ}} \cdot \frac{\sin 66^{\circ}}{\cos 66^{\circ}} \cdot \frac{\sin 78^{\circ}}{\cos 78^{\circ}} \\
& =\frac{\left(2 \sin 66^{\circ} \sin 6^{\circ}\right)\left(2 \sin 78^{\circ} \sin 42^{\circ}\right)}{\left(2 \cos 66^{\circ} \cos 6^{\circ}\right)\left(2 \cos 78^{\circ} \cos 42^{\circ}\right)} \\
& =\frac{\cos \left(66^{\circ}-6^{\circ}\right)-\cos \left(66^{\circ}+6^{\circ}\right)}{\cos \left(66^{\circ}+6^{\circ}\right)+\cos \left(66^{\circ}-6^{\circ}\right)} \\
& . \frac{\cos \left(78^{\circ}-42^{\circ}\right)-\cos \left(78^{\circ}+42^{\circ}\right)}{\cos \left(78^{\circ}+42^{\circ}\right)+\cos \left(78^{\circ}-42^{\circ}\right)} \\
& =\frac{\left(\cos 60^{\circ}-\cos 72^{\circ}\right)\left(\cos 36^{\circ}-\cos 120^{\circ}\right)}{\left(\cos 60^{\circ}+\cos 72^{\circ}\right)\left(\cos 36^{\circ}+\cos 120^{\circ}\right)} \\
& =\frac{\left(\cos 60^{\circ}-\sin 18^{\circ}\right)\left(\cos 36^{\circ}+\sin 30^{\circ}\right)}{\left(\cos 60^{\circ}+\sin 18^{\circ}\right)\left(\cos 36^{\circ}-\sin 30^{\circ}\right)} \\
& \ldots\left[\because \cos \left(90^{\circ}+\theta\right)=-\sin \theta\right] \\
& =\frac{\left(\frac{1}{2}-\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}+\frac{1}{2}\right)}{\left(\frac{1}{2}+\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}-\frac{1}{2}\right)} \\
& =\frac{9-5}{5-1}\\
& =1 \\
& =\text { R.H.S. } \\
\end{aligned}