Solve tan 10 - tan 50 +tan 70 with trigonometry.

Below is the solution, I hope it helps.

  i) tan(70) - tan(50) = tan(60 + 10) - tan(60 - 10) 

= {tan(60) + tan(10)}/{1 - tan(60)*tan(10)} - {tan(60) - tan(10)}/{1 + tan(10)*tan(60)} 

ii) Taking LCM & simplifying with applying tan(60) = √3, the above simplifies to: 

= 8*tan(10)/{1 - 3*tan²(10)} 

iii) So tan(70) - tan(50) + tan(10) = 8*tan(10)/{1 - 3*tan²(10)} + tan(10) 

= [8*tan(10) + tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)} 

= [9*tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)} 

= 3 [3*tan(10) - tan³(10)]/{1 - 3*tan²(10)} 

= 3*tan(30) = 3*(1/√3) = √3 [Proved] 

[Since tan(3A) = {3*tan(A) - tan³(A)}/{1 - 3*tan²(A)}, 
{3*tan(10) - tan³(10)}/{1 - 3*tan²(10)} = tan(3*10) = tan(30)]

Y=3x+1 so the first one