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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
Prove that:
$\tan20^\circ\tan30^\circ\tan40^\circ\tan80^\circ=1$
✓

Answer

$\text{LHS}=\tan20^\circ\tan30^\circ\tan40^\circ\tan80^\circ$
$\frac{1}{\sqrt3}(\tan20^\circ\tan40^\circ\tan80^\circ)$$\Big[\because\ \tan30^\circ=\frac{1}{\sqrt3}\Big]$
$=\ \frac{(\sin20^\circ\sin40^\circ\sin80^\circ)}{(\cos20^\circ\cos40^\circ\cos80^\circ)\sqrt3}$
$=\ \frac{(2\sin20^\circ\sin40^\circ)\sin80^\circ}{\sqrt3(2\cos20^\circ\cos40^\circ)\cos80^\circ}$
Applying
$\Rightarrow\ 2\sin\text{A}\sin\text{B}=\cos(\text{A}-\text{B})-\cos(\text{A+B})$
$2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})$
$=\ \frac{(\cos(40^\circ-20^\circ)-\cos(20^\circ+40^\circ))\sin80^\circ}{\cos(20^\circ+40^\circ)+\cos(40^\circ-20^\circ)\cos80^\circ\sqrt3}$
$=\ \frac{(\cos20^\circ-\cos60^\circ)\sin80^\circ}{\sqrt3(\cos60^\circ+\cos20^\circ)\cos80^\circ}$
$=\ \frac{\Big(\cos20^\circ-\frac{1}{2}\Big)\sin80^\circ}{\sqrt3\Big(\frac{1}{2}+\cos20^\circ\Big)\cos80^\circ}$
$=\ \frac{2\sin20^\circ\sin80^\circ-\sin80^\circ}{\sqrt3(\cos80^\circ+2\cos20^\circ\cos80^\circ)}$
Now,
$\Rightarrow\ 2\sin\text{A}\cos\text{B}=\sin(\text{A+B})+\sin(\text{A}-\text{B})$
$=\ \frac{\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ}{\sqrt3(\cos80^\circ+\cos(20^\circ+80^\circ)+\cos(80^\circ-20^\circ))}$
$=\ \frac{\sin100^\circ+\sin60^\circ-\sin80^\circ}{\sqrt3(\cos80^\circ+\cos100^\circ+\cos60^\circ)}$
$=\ \frac{\sin100^\circ+\sin60^\circ-\sin(80^\circ-100^\circ)}{\sqrt3(\cos80^\circ+\cos(1800^\circ-80^\circ)+\sin60^\circ)}$
$=\ \frac{\sin100^\circ+\frac{\sqrt3}{2}-\sin100^\circ}{\sqrt3(\cos80^\circ-\cos80^\circ+\cos60^\circ)}$
$=\ \frac{\frac{\sqrt3}{2}}{\sqrt3\big(\frac{1}{2}\big)}=1=\ \text{RHS}$

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