Prove that :$\tan 70 - \tan 20 = 2\tan 40 + 4\tan 10$ My Attempt,  $$70-20=40+10$$ $$\tan (70-20)=\tan (40+10)$$ $$\dfrac {\tan 70 - \tan 20}{1+\tan 70. \tan 20 }=\dfrac  {\tan 40 + \tan 10 }{1-\tan 40. \tan 10 }$$   How should I move on?  Please help   Thanks
I am currently studying in class 10 and I am unable to do this problem. $$\tan 70 ° -\tan 20° -2 \tan 40° =4\tan 10°$$ Can anybody please help me.  Thanks!
$$\tan(\theta_1+\theta_2+\cdots+\theta_n)=\frac{S_1-S_3+S_5+\cdots}{1-S_2+S_4+\cdots}$$  where  $S_i$ denotes the sum of product of tangent of angles taken $i$ at a time.     For example,$$\tan(\theta_1+\theta_2)=\frac{S_1}{1-S_2}=\frac{\tan\theta_1+\tan\theta_2} {1-\tan\theta_1\tan\theta_2}$$   (This formula is given  in my textbook with no derivation or background)  How to derive this?