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Given : (arctanx)^2 + (arccotx)^2 = 5π^2/8 Find x. Please solve it as soon as possible

Given : (arctanx)^2 + (arccotx)^2 = 5π^2/8 Find x. Please solve it as soon as possible

Grade:12

1 Answers

Ajay
209 Points
4 years ago
Some how answer got deleted. Posting again----------------------------------------------------------------------------------------
We\quad Know\quad that\quad { cot }^{ -1 }x\quad =\quad \pi /2\quad -{ tan }^{ -1 }x\\ substituting\quad in\quad orignal\quad equation\quad it\quad becomes\\ { { (tan }^{ -1 }x) }^{ 2 }\quad +\quad { (\pi /2\quad -{ tan }^{ -1 }x) }^{ 2 }\quad =\quad 5{ \pi }^{ 2 }/8\\ On\quad simplifying\quad the\quad equation\quad becomes\\ { { 16(tan }^{ -1 }x) }^{ 2 }\quad { -8\pi { tan }^{ -1 }x }\quad -3{ \pi }^{ 2 }\quad =\quad 0.\\ (4{ tan }^{ -1 }x-3\pi )(4{ tan }^{ -1 }x+\pi )\quad =\quad 0\\ { tan }^{ -1 }x\quad =\quad 3\pi /4\quad or\quad { tan }^{ -1 }x\quad =\quad -\pi /4\\ x\quad =\quad tan3\pi /4\quad or\quad \quad x\quad =\quad tan(-\pi /4)\quad \\ in\quad both\quad cases\quad x\quad \quad =\quad -1\quad is\quad the\quad answer\quad

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