# Proving $\tan^2 20^{\circ} + \frac{3}{16}\csc^2 40^ {\circ}\sec^2 20^{\circ}-\frac{\sqrt3}{4}\tan 20^{\circ} \sec^2 20^{\circ} = 4 \sin^2 20^{\circ}$

How to prove that $$\tan^2 20^{\circ} + \frac{3}{16}\csc^2 40^ {\circ}\sec^2 20^{\circ}-\frac{\sqrt3}{4}\tan 20^{\circ} \sec^2 20^{\circ} = 4 \sin^2 20^{\circ}$$

Can I express them in terms of $\tan 20^{\circ}$ and prove it?

• prove the identity, that means prove that left hand side equals right hand side – Ray Cheng Jun 6 '17 at 5:01
• Are these all degrees or radian? – Jaideep Khare Jun 6 '17 at 5:04
• all degrees, I am sure – Ray Cheng Jun 6 '17 at 5:04
• What is the source? – lab bhattacharjee Jun 6 '17 at 5:14
• What have you tried. I find it easiest to put this in terms for sins and cosines. A lot should simplify. the only tricky part I see is $\csc^2 40 = \frac 1{\sin (20 + 20)}$ which should be straight forward. – fleablood Jun 6 '17 at 5:21

Let $x=20^\circ$. Then \begin{eqnarray} &&\tan^2 20^{\circ} + \frac{3}{16}\csc^2 40^ {\circ}\sec^2 20^{\circ}-\frac{\sqrt3}{4}\tan 20^{\circ} \sec^2 20^{\circ}-4\sin^2 20^\circ\\ &=&\frac{1}{\cos^2 x\sin^2(2x)}\bigg[\frac{3}{16}-\sqrt3\cos x\sin^3x+4\cos^2x\sin^4x-\sin^4(2x)\bigg].\\ \end{eqnarray} Noting $$\sin 60^\circ=\sin(3x)=-4\sin^3x+3\sin x=\frac{\sqrt3}{2}, \cos 60^\circ=\cos(3x)=4\cos^3x-3\cos x=\frac{1}{2}$$ and $$\sin 120^\circ=\sin(6x)=-4\sin^3(2x)+3\sin(2x)=\frac{\sqrt3}{2}$$ one has $$4\sin^3x=-\frac{\sqrt3}{2}+3\sin x,4\cos^3x=\frac12+3\cos x, 4\sin^3(2x)=-\frac{\sqrt3}{2}+3\sin(2x)$$ and hence \begin{eqnarray} &&\frac{3}{16}-\sqrt3\cos x\sin^3x+4\cos^2x\sin^4x-\sin^4(2x)\\ &=&\frac{3}{16}-\frac{\sqrt3}{4}\cos x\bigg(-\frac{\sqrt3}{2}+3\sin x\bigg)+\cos^2 x\sin x\bigg(-\frac{\sqrt3}{2}+3\sin x\bigg)\\ &&-\frac14\sin(2x)\bigg(-\frac{\sqrt3}{2}+3\sin(2x)\bigg)\\ &=&\frac{3}{16}+\frac{3}{8}\cos x-\frac{\sqrt{3}}{4}\sin (2x)-\frac{\sqrt{3}}{4}\cos x \sin(2x)\\ &=&\frac{3}{16}+\frac{3}{8}\cos x-\frac{\sqrt{3}}{4}\sin (2x)-\frac{\sqrt{3}}{8}(\sin 60°+\sin x)\\ &=&\frac18 \bigg(3\cos x-\sqrt3 \bigg(\sin x+2\sin(2x)\bigg)\bigg). \end{eqnarray} Noting $$\sin x+2\sin(2x)=\sin x+2\sin(60^\circ-x)=\sqrt3\cos x$$ it is easy to check that $$3\cos x-\sqrt3(\sin x+2\sin(2x))=0$$ and hence $$\frac{3}{16}-\sqrt3\cos x\sin^3x+4\cos^2x\sin^4x-\sin^4(2x)=0.$$
After I asked a question at Quora, I am sure that in order to show that $3\cos x-\sqrt3(\sin x+2\sin(2x))=0$, where $x=20°$, this is actually equivalent to show that $\tan 60°\cos20°-\sin20°=2\sin 40°$. Using this approach, we don't need to use imaginary numbers (Euler's formula).
$\tan 60°\cos20°-\sin20°=\frac{1}{\cos 60°}(\sin 60° \cos 20°-\cos 60° \sin 20°)=2 \sin(60°-20°)=2 \sin 40°.$