# Find the $A$ :$A=\frac{\tan 30^{\circ}+\tan 40^{\circ}+\tan 50^{\circ}+\tan 60^{\circ}}{\cos20^{\circ}}=?$

Find the $A$ :

$$A=\frac{\tan 30^{\circ}+\tan 40^{\circ}+\tan 50^{\circ}+\tan 60^{\circ}}{\cos20^{\circ}}=?$$

My Try :

$$\tan q+\tan q=\frac{\sin (q+p)}{\cos q\cos p}\\\cos q \cos p=\frac{\cos (q+p)+\cos(q-p)}{2}\\A=\frac{\frac{\sin(90^{\circ})}{\cos 40^{\circ} \cos 50}+\frac{\sin(90^{\circ})}{\cos 30^{\circ} \cos 40^{\circ}}}{\cos 20^{\circ}}=\frac{\cos 30^{\circ} \cos 40^{\circ} +\cos 40^{\circ} \cos 50^{\circ}}{\cos 30^{\circ}\cos 60^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 20^{\circ} }$$

now ?

Generalization: $$\tan(2a-3d)+\tan(2a-d)+\tan(2a+d)+\tan(2a+3d)$$

$$=\dfrac{\sin(2a-3d+2a+3d)}{\cos(2a-3d)\cos(2a+3d)}+\dfrac{\sin(2a-d+2a+d)}{\cos(2a-d)\cos(2a+d)}$$

$$=\dfrac{\sin4a\{2\cos(2a-3d)\cos(2a+3d)+2\cos(2a-d)\cos(2a+d)\}}{2\cos(2a-3d)\cos(2a-d)\cos(2a+d)\cos(2a+3d)}$$

$$=\dfrac{\sin4a(2\cos4a+\cos6d+\cos2d)}{2\cos(2a-3d)\cos(2a-d)\cos(2a+d)\cos(2a+3d)}$$

If $4a=90^\circ,2a+3d=90^\circ-(2a-3d);2a+d=90^\circ-(2a-d)$

$$\tan(45^\circ-3d)+\tan(45^\circ-d)+\tan(45^\circ+d)+\tan(45^\circ+3d)$$

$$=\dfrac{\cos6d+\cos2d}{2\cos(2a-3d)\cos(2a-d)\cos(2a+d)\cos(2a+3d)}$$

$$=\dfrac{4\cos4d\cos2d}{2\cos(2a-3d)\sin(2a-3d)\cdot2\sin(2a-d)\cos(2a-d)}$$

$$=\dfrac{4\cos4d\cos2d}{\sin(4a-6d)\sin(4a-2d)}$$

$$=\dfrac{4\cos4d\cos2d}{\cos6d\cos2d}\text{ as }4a=90^\circ$$

$$=\dfrac{4\cos4d}{\cos6d}\text{ for }\cos2d\ne0$$

Here $d=5^\circ$

$$\tan40^{\circ}+\tan50^{\circ}=\dfrac1{\cos40^{\circ}\cos50^{\circ}}=\dfrac1{\cos40^{\circ}\sin40^{\circ}}=\dfrac2{\sin(2\cdot40^{\circ})}=\dfrac2{\cos10^{\circ}}$$

$$\tan30+\tan60=\dfrac4{\sqrt3}$$

So we need $$\dfrac{\dfrac2{\cos10^{\circ}}+\dfrac4{\sqrt3}}{\cos20^{\circ}}=4\cdot\dfrac{\dfrac{\sqrt3}2+\cos10^{\circ}}{\sqrt3\cos20^{\circ}\cos10^{\circ}}$$

Now $\dfrac{\sqrt3}2+\cos10^{\circ}=\cos30^{\circ}+\cos10^{\circ}=2\cos20^{\circ}\cos10^{\circ}$

Can you take it from here?