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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae4 Marks
Question
Prove that:
$\tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ=3$
✓

Answer

$\text{LHS}=\tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ$
$=\ (\tan20^\circ\tan40^\circ\tan80^\circ)\sqrt3$$[\because\ \tan60^\circ=\sqrt3]$
$=\ \Big(\frac{\sin20^\circ\sin40^\circ\sin80^\circ}{\cos20^\circ\cos40^\circ\cos80^\circ}\Big)\sqrt3$
$=\ \frac{(2\sin20^\circ\sin40^\circ)\sin80^\circ\times\sqrt3}{(2\cos20^\circ\cos40^\circ)\cos80^\circ}$
Applying
$\Rightarrow\ 2\sin\text{A}\sin\text{B}-\cos(\text{A}-\text{B})-\cos(\text{A+B})$
$2\cos\text{A}\cos\text{B}-\cos(\text{A+B})+\cos(\text{A}-\text{B})$
$=\ \frac{(\cos(40^\circ-20^\circ)-\cos(40^\circ+20^\circ))\sin80^\circ\times\sqrt3}{(\cos(20^\circ+40^\circ)+\cos(40^\circ-20^\circ))\cos80^\circ}$
$=\ \frac{(\cos20^\circ-\cos60^\circ)\sin80^\circ\times\sqrt3}{(\cos60^\circ+\cos20^\circ)\cos80^\circ}$
$=\ \frac{\Big(\cos20^\circ-\frac{1}{2}\Big)\sin80^\circ\times\sqrt3}{\Big(\frac{1}{2}+\cos20^\circ\Big)\cos80^\circ}$
$=\ \frac{(2\sin80^\circ\cos20^\circ-\sin80^\circ)\sqrt3}{\cos80^\circ+2\cos20^\circ\cos80^\circ}$
$\Rightarrow\ 2\sin\text{A}\cos\text{B}-\sin(\text{A+B})+\sin(\text{A}-\text{B})$
$=\ \frac{(\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ)\sqrt3}{\cos80^\circ+(\cos(20^\circ+80^\circ)+\cos(80^\circ-20^\circ))}$
$=\ \frac{(\sin100^\circ+\sin60^\circ-\sin80^\circ)\sqrt3}{\cos80^\circ+\cos(180^\circ-80^\circ)+\cos60^\circ}$
$=\ \frac{\Big(\sin(180^\circ+\frac{\sqrt3}{2}-\sin80^\circ)\Big)\sqrt3}{\cos80^\circ-\cos80^\circ+\cos60^\circ}$
$=\ \frac{\frac{3}{2}}{\frac{1}{2}}=3=\text{RHS}$

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