What is the integral value of $\frac{\tan 20^\circ+\tan40^\circ+\tan80^\circ-\tan60^\circ}{\sin40^\circ}$?

Question

I have tried possibly all approaches.

I first expressed $80$ as $60+20$ and $40$ as $60-20$ and then used trig identities.I later used conditional identities expressing $\tan 20^\circ+\tan40^\circ+\tan120^\circ$ as $\tan 20^\circ \tan40^\circ \tan120^\circ$. But I really can't get to the end of it .

Please help.

Answer 1

$$\tan20^\circ-\tan60^\circ=-\dfrac{\sin(60-20)^\circ}{\cos20^\circ\cdot\cos60^\circ}=-\dfrac{2\sin40^\circ}{\cos20^\circ}$$

$$\tan40^\circ+\tan80^\circ=\dfrac{\sin(40+80)^\circ}{\cos40^\circ\cos80^\circ}$$

Adding $(1),(2)$

$$\dfrac{\sin120^\circ}{\cos40^\circ\cos80^\circ}-\dfrac{2\sin40^\circ}{\cos20^\circ} =\dfrac{\sin120^\circ\cos20^\circ-2\sin40^\circ\cos40^\circ\cos80^\circ} {\cos20^\circ\cos40^\circ\cos80^\circ}$$

Now $S=\sin120^\circ\cos20^\circ-2\sin40^\circ\cos40^\circ\cos80^\circ$

$2S=\sin(120+20)^\circ+\sin(120-20)^\circ-2\sin80^\circ\cos80^\circ$

$=\sin(180-40)^\circ+\sin100^\circ-\sin160^\circ$

$=\sin40^\circ+\sin80^\circ-\sin20^\circ$

$=\sin40^\circ+2\sin30^\circ\cos50^\circ$

$=2\sin40^\circ$

Formulas used :

• $\sin(180^\circ-A)=\sin A$

• Prosthaphaeresis Formula $:\sin C-\sin D$

• $\sin2y=2\sin y\cos y$

• $2\sin A\cos B=\sin(A+B)+\sin(A-B)$

Now use Upon multiplying $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ by the sine of a certain angle, it gets reduced. What is that angle? to find the answer to be $$\dfrac1{\dfrac18}=?$$

Answer 2

Using $\tan20^\circ\cdot\tan40^\circ\cdot\tan80^\circ=\tan60^\circ$ (Proof)

$$\tan20^\circ+\tan40^\circ+\tan80^\circ-\tan60^\circ$$

$$=\tan20^\circ+\tan40^\circ+\tan80^\circ-\tan20^\circ\cdot\tan40^\circ\cdot\tan80^\circ$$

$$=\tan20^\circ(1-\tan40^\circ\cdot\tan80^\circ)+\tan40^\circ+\tan80^\circ$$

$$=(1-\tan40^\circ\cdot\tan80^\circ)\left(\tan20^\circ+\dfrac{\tan40^\circ+\tan80^\circ}{1-\tan40^\circ\cdot\tan80^\circ}\right)$$

$$=\dfrac{\cos(40^\circ+80^\circ)}{\cos40^\circ\cos80^\circ}\left(\tan20^\circ+\tan(40^\circ+80^\circ)\right)$$

$$=\dfrac{\cos120^\circ}{\cos40^\circ\cos80^\circ}\cdot\dfrac{\sin(20^\circ+120^\circ)}{\cos20^\circ\cdot\cos120^\circ}$$

$$=\dfrac{\sin40^\circ}{\cos20^\circ\cos40^\circ\cos80^\circ}\text{ Using }\sin(180^\circ-A)=\sin A$$

Now use Upon multiplying $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ by the sine of a certain angle, it gets reduced. What is that angle?

Answer 3

Using the comment by Roman83, $$\tan20^\circ+\tan80^\circ−\tan40^\circ=3\tan60^\circ$$(Proof is in Method$\#2$ here)

the numerator can be reduced to

$$2(\tan60^\circ+\tan40^\circ)=\dfrac{2\sin(60+40)^\circ}{\cos60^\circ\cdot\cos40^\circ}$$

Using $\sin(180^\circ-A)=\sin A,\sin100^\circ=\sin80^\circ$

Using $\sin2B=2\sin B\cos B,\sin80^\circ=2\sin40^\circ\cos40^\circ$

and we are done!

Answer 4

First, we can write the expression in terms of only $\sin$ and $\cos$,

$$\tan 20^\circ + \tan 40^\circ + \tan 80^\circ - \tan 60^\circ$$ $$= \frac{\sin 20^\circ}{\cos 20^\circ} + \frac{\sin 40^\circ}{\cos 40^\circ} + \frac{\sin 80^\circ}{\cos 80^\circ} - \sqrt{3}$$ $$= \frac{\sin 20^\circ \cos 40^\circ \cos 80^\circ + \cos 20^\circ \sin 40^\circ \cos 80^\circ + \cos 20^\circ \cos 40^\circ \sin 80^\circ}{\cos 20^\circ \cos 40^\circ \cos 80^\circ} - \sqrt{3}$$

Since $\cos \theta \cdot \cos \left(60^\circ - \theta \right) \cdot \cos \left(60^\circ + \theta \right) = \frac{1}{4} \cos 3\theta$, we get,

$$\frac{2\left[2 \sin 20^\circ \cos 40^\circ \cos 80^\circ + 2 \cos 20^\circ \sin 40^\circ \cos 80^\circ + 2 \cos 20^\circ \cos 40^\circ \sin 80^\circ \right]}{\cos 60^\circ} - \sqrt{3}$$

Pairing terms in the numerator we get,

$$4\left[\cos 20^\circ \left(\sin 40^\circ \cos 80^\circ + \cos 40^\circ \sin 80^\circ \right) + \cos 40^\circ \left(\sin 20^\circ \cos 80^\circ + \cos 20^\circ \sin 80^\circ \right) + \cos 80^\circ \left(\sin 20^\circ \cos 40^\circ + \cos 20^\circ \sin 40^\circ \right) \right] - \sqrt{3}$$

Since $\sin(A + B) = \sin A \cos B + \cos A \sin B$, we get,

$$4\left[\cos 20^\circ \sin 120^\circ + \cos 40^\circ \sin 100^\circ + \cos 80^\circ \sin 60^\circ \right] - \sqrt{3}$$ $$= 4\left[\frac{\sqrt{3}}{2} \left(\cos 20^\circ + \cos 80^\circ \right) + \cos 40^\circ \sin 100^\circ \right] - \sqrt{3}$$

Since $\cos A + \cos B = 2\cos(\frac{A + B}{2})\cos(\frac{A - B}{2})$ and $\cos A \sin B = \sin(B + A) + \sin(B - A)$, we get,

$$4\left[\sqrt{3} \cos 50^\circ \cos 30^\circ + \frac{1}{2} \left(\sin 140^\circ + \sin 60^\circ \right) \right] - \sqrt{3}$$

Since $\cos(90^\circ - \theta) = \sin \theta$ and $\sin(180^\circ - \theta) = \sin \theta$, we get,

$$4\left[2\sin 40^\circ + \frac{\sqrt{3}}{4} \right] - \sqrt{3}$$ $$= \boxed{8\sin 40^\circ}$$