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Question
Prove that `2 tan^-1 (1/8) + tan^-1 (1/7) + 2tan^-1 (1/5) = pi/4`
Sum
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Solution
L.H.S. = `2 tan^-1 (1/8) + tan^-1 (1/7) + 2 tan^-1 (1/5)`
= `2[tan^-1 (1/8) + tan^-1 (1/5)] + tan^-1 (1/7)`
= `2[tan^-1 ((1/8 + 1/5)/(1 - 1/8 xx 1/5))] + tan^-1 (1/7)`
= `2[tan^-1 ((13/40)/(39/40))] + tan^-1 (1/7)`
= `2tan^-1 (1/3) + tan^-1 (1/7)`
= `tan^-1 (1/3) + tan^-1 (1/3) + tan^-1 (1/7)`
= `tan^-1 ((1/3 + 1/3)/(1 - 1/3 xx 1/3)) + tan^-1 (1/7)`
= `tan^-1 ((2/3)/(8/9)) + tan^-1 (1/7)`
= `tan^-1 (3/4) + tan^-1 (1/7)`
= `tan^-1 ((3/4 + 1/7)/(1 - 3/4 xx 1/7))`
= `tan^-1 ((25/28)/(25/28))`
= `tan^-1 (1)`
= `pi/4`
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