Explanation Given information: The inequality, tan − 1 ( 1 ) + tan − 1 ( 2 ) + tan − 1 ( 3 ) = 180 ∘ Suppose that, A B ^ C = A C ^ B = B E ^ D = π 2 E D ^ F = θ 1 , B D ^ E = θ 2 , B D ^ C = θ 3 θ 1 + θ 2 + θ 3 = π Formula used: tan ( A + B ) = tan ( A ) + tan ( B ) 1 − tan ( A ) ⋅ tan ( B ) . Proof: By the B C F Δ , D F ^ E = π 2 − θ 1 Therefore, tan ( π 2 − θ 1 ) = B C C F = 3 3 = 1 cot θ 1 = 1 tan θ 1 = 1 θ 1 = tan − 1 ( 1 ) By the B C D Δ

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 1.5, Problem 56PS

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To prove: that tan−1(1)+tan−1(2)+tan−1(3)=180∘ .

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Chapter 1.5, Problem 55PS

Chapter 1.5, Problem 57PS

Chapter 1 Solutions

Calculus

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To prove: that tan − 1 ( 1 ) + tan − 1 ( 2 ) + tan − 1 ( 3 ) = 180 ∘ .

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To prove: that tan−1(1)+tan−1(2)+tan−1(3)=180∘ .

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Chapter 1 Solutions