Question

Solve this:
Q.6. Show that :
tan 50$°$ = tan 40$°$ + 2 tan 10$°$

Answer 1

$$\begin{gathered} Weknow \\ \tan \left(x-y\right)=\frac{\tan x-\tan y}{1+\tan x\tan y} \\ \tan \left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x\tan y} \\ \tan 10°=\tan \left(50°-40°\right) \\ \tan 10°=\frac{\tan 50°-\tan 40°}{1+\tan 50°\tan 40°}\_\_\_\_\_\_\_\_\_\left(1\right) \\ Now \\ \tan 90°=\tan \left(50°+40°\right) \\ \tan 90°=\frac{\tan 50°+\tan 40°}{1-\tan 50°\tan 40°} \\ But\tan 90°\to \infty ,thismeansdenominatorin\frac{\tan 50°+\tan 40°}{1-\tan 50°\tan 40°}mustbe0 \\ 1-\tan 50°\tan 40°=0 \\ \tan 50°\tan 40°=1 \\ Putitinequation1 \\ \tan 10°=\frac{\tan 50°-\tan 40°}{1+1} \\ \tan 10°=\frac{\tan 50°-\tan 40°}{2} \\ 2\tan 10°=\tan 50°-\tan 40° \\ \tan 50°=2\tan 10°+\tan 40° \end{gathered}$$