Prove that arctan x-arctan y-n-arctanx-y-1-x y if x-0- y-0 and x y-1 - And arctan x-arctan y-arctanx-y-1-x y-n- if x1

If x gt; 0, y gt; 0 such that xy gt; 1, then ( x+y/1 xy ) is positive and therefore, ( x+y/1 xy ) is positive angle between 0^∘ and 90^∘.. Similarly, ...

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Standard XII

Mathematics

Basic Inverse Trigonometric Functions

Prove that ar...

Question

Prove that arctan(x) + arctan(y) = n + arctan $(\frac{x + y}{1 - x y})$ if x > 0, y > 0 and xy > 1. And arctan(x) + arctan(y) = arctan
$(\frac{x + y}{1 - x y}) - n$ , if x < 0, y < 0 and xy > 1.

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Solution

If x > 0, y > 0 such that xy > 1, then $(\frac{x + y}{1 - x y})$ is positive and therefore, $(\frac{x + y}{1 - x y})$ is positive angle between $0^{\circ}$ and $90^{\circ}$ ..
Similarly, if x < 0, y < 0 such that xy > 1, then $\frac{x + y}{1 - x y}$ is positive and therefore, $t a n^{- 1} (\frac{x + y}{1 - x y})$ is a negative angle while $t a n^{- 1} x + t a n^{- 1} y$ is a positive angle while $t a n^{- 1} x + t a n^{- 1} y$ is a non-negative angle. Therefore, $t a n^{- 1} x + t a n^{- 1} y = n + t a n^{- 1} (\frac{x + y}{1 - x y})$ , if x > 0, y > 0 and xy > 1 and arctan(x) + arctan(y) = arctan $(\frac{x + y}{1 - x y}) - n$ , if x < 0, y < 0 and xy > 1.

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Prove that arctan(x) + arctan(y) = n + arctan (x+y1−xy) if x > 0, y > 0 and xy > 1. And arctan(x) + arctan(y) = arctan (x+y1−xy)−n, if x < 0, y < 0 and xy > 1.

Prove that arctan(x) + arctan(y) = n + arctan $(\frac{x + y}{1 - x y})$ if x > 0, y > 0 and xy > 1. And arctan(x) + arctan(y) = arctan
$(\frac{x + y}{1 - x y}) - n$ , if x < 0, y < 0 and xy > 1.