Prove that: tan−1(21)+tan−1(51)+tan−1(81)=4π

=tan−1(21)+tan−1(51)+tan−1(81)=tan−11−21×5121+51+tan−1(81) [∵21×51=101<1]=tan−1(97)+tan−1(81)=tan−1⎝⎛1−97×8197+81⎠⎞=tan−1(7265×6572)=tan−1(1)=4π=RHS

Get 2 FREE Instant-Explanations on Filo with code FILOAPP

Prove that: ^{-1} ({1}{2}) + ^{-1} ({1}{5}) + ^{-1} ({1}{8}) =

Question

Prove that: $tan^{- 1} (\frac{1}{2}) + tan^{- 1} (\frac{1}{5}) + tan^{- 1} (\frac{1}{8}) = \frac{π}{4}$

Views: 812 students

Found 3 tutors discussing this question

Levi Discussed

Prove that:

tan^{- 1} (\frac{1}{2}) + tan^{- 1} (\frac{1}{5}) + tan^{- 1} (\frac{1}{8}) = \frac{π}{4}

11 mins ago

Discuss this question LIVE

11 mins ago

= tan^{- 1} (\frac{1}{2}) + tan^{- 1} (\frac{1}{5}) + tan^{- 1} (\frac{1}{8})

= tan^{- 1} \frac{\frac{1}{2} + \frac{1}{5}}{1 - \frac{1}{2} \times \frac{1}{5}} + tan^{- 1} (\frac{1}{8})

[∵ \frac{1}{2} \times \frac{1}{5} = \frac{1}{10} < 1]

= tan^{- 1} (\frac{7}{9}) + tan^{- 1} (\frac{1}{8}) = tan^{- 1} ⎝ ⎛ \frac{\frac{7}{9} + \frac{1}{8}}{1 - \frac{7}{9} \times \frac{1}{8}} ⎠ ⎞ = tan^{- 1} (\frac{65}{72} \times \frac{72}{65})

= tan^{- 1} (1) = \frac{π}{4} = R H S

Was this solution helpful?

150

Report

One destination to cover all your homework and assignment needs

Learn Practice Revision Succeed

Instant 1:1 help, 24x7

60, 000+ Expert tutors

Textbook solutions

Big idea maths, McGraw-Hill Education etc

Essay review

Get expert feedback on your essay

Schedule classes

High dosage tutoring from Dedicated 3 experts

Trusted by 4 million+ students

Stuck on the question or explanation?

Connect with our maths tutors online and get step by step solution of this question.

231 students are taking LIVE classes

Question Text	Prove that: $tan^{- 1} (\frac{1}{2}) + tan^{- 1} (\frac{1}{5}) + tan^{- 1} (\frac{1}{8}) = \frac{π}{4}$
Answer Type	Text solution:1
Upvotes	150