prove that tan 70 -tan50 +tan10 =tan60

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Solution

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)}

Taking LCM & simplifying with applying tan(60) = âˆš3, the above simplifies to:

= 8tan(10)/{1 - 3tan²(10)}

Consider
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/âˆš3) = âˆš3 [Proved]

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

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