How do you prove $cot^-1(x)=pi/2-tan^-1(x)$ ?

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Answer 1 · 2016-07-18T05:03:31

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The solution

Let $alpha$ be the angle with tangent $=x$
Let $beta=pi/2-alpha$ be the angle with cotangent $=x$

Then $alpha+beta=pi/2$

$cot^-1 x=pi/2-tan^-1 x$

$cot^-1 x=pi/2-(pi/2-cot^-1 x)$

$cot^-1 x=pi/2-pi/2+cot^-1 x$

$cot^-1 x=cot^-1 x$

God bless....I hope the explanation is useful.

How do you prove cot^-1(x)=pi/2-tan^-1(x)cot^-1(x)=pi/2-tan^-1(x)?