{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:
= 8tan(10)/{1 - 3tan²(10)}
iii) So tan(70) - tan(50) + tan(10) = 8tan(10)/{1 - 3tan²(10)} + tan(10)
= [8tan(10) + tan(10) - 3tan³(10)]/{1 - 3tan²(10)}
= [9tan(10) - 3tan³(10)]/{1 - 3tan²(10)}
= 3 [3tan(10) - tan³(10)]/{1 - 3tan²(10)}
= 3tan(30) = 3(1/√3) = √3 [Proved]
[Since tan(3A) = {3tan(A) - tan³(A)}/{1 - 3tan²(A)},
{3tan(10) - tan³(10)}/{1 - 3tan²(10)} = tan(3*10) = tan(30)]
