tan⁻¹ 1/5 + tan⁻¹ 1/7 + tan⁻¹ 1/3 + tan⁻¹ 1/8=π/4,Prove it.
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In the attachment I have answered this problem. Concept: (tan inverse of x) +(tan inverse of y) = tan inverse of ((x+y)/(1-xy)) See the attachment for detailed solution.
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LHS = tan^-1(1/5) + tan^-1(1/7) + tan^-1(1/3) + tan^-1(1/8)
we know, tan^-1x + tan^-1y = tan^-1(x + y)/(1 - xy)
where , xy < 1
= [tan^-1(1/5) + tan^-1(1/7) ] + [tan^-1(1/3) + tan^-1(1/8) ]
= tan^-1{(1/5 + 1/7)/(1 - 1/5 × 1/7)} + tan^-1{(1/3 + 1/8)/(1 - 1/3 × 1/8)}
= tan^-1{(12/35}/(34/35)} + tan^-1{(11/24)/(23/24}
= tan^-1{6/17} + tan^-1{11/23}
= tan^-1{(6 × 23 + 17 × 11)/17 × 23}/{1 - 6/17 × 11/23}
= tan^-1{(138 + 187)/(391 - 66)}
= tan^-1{325/325}
= tan^-1(1) = π/4 = RHS
we know, tan^-1x + tan^-1y = tan^-1(x + y)/(1 - xy)
where , xy < 1
= [tan^-1(1/5) + tan^-1(1/7) ] + [tan^-1(1/3) + tan^-1(1/8) ]
= tan^-1{(1/5 + 1/7)/(1 - 1/5 × 1/7)} + tan^-1{(1/3 + 1/8)/(1 - 1/3 × 1/8)}
= tan^-1{(12/35}/(34/35)} + tan^-1{(11/24)/(23/24}
= tan^-1{6/17} + tan^-1{11/23}
= tan^-1{(6 × 23 + 17 × 11)/17 × 23}/{1 - 6/17 × 11/23}
= tan^-1{(138 + 187)/(391 - 66)}
= tan^-1{325/325}
= tan^-1(1) = π/4 = RHS
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