PROVE
arctan x + arctan y = π + arctan (x+y/1-xy) , x,y>0 and xy >1
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Hi friend
here is proof to arctan x + arctan y = π + arctan (x+y/1-xy) , x,y>0 and xy >1
If x > 0, y > 0 such that xy > 1, then x+y1−xyx+y1−xy is positive and therefore, x+y1−xyx+y1−xy is positive angle between 0° and 90°.
Similarly, if x < 0, y < 0 such that xy > 1, then x+y1−xyx+y1−xy is positive and therefore, tan−1−1 (x+y1−xyx+y1−xy) is a negative angle while tan−1−1 x + tan−1−1 y is a positive angle while tan−1−1 x + tan−1−1 y is a non-negative angle. Therefore, tan−1−1 x + tan−1−1 y = π + tan−1−1 (x+y1−xyx+y1−xy), if x > 0, y > 0 and xy > 1 and
arctan(x) + arctan(y) = arctan(x+y1−xyx+y1−xy) - π, if x < 0, y < 0 and xy > 1.
hope this is helpful
here is proof to arctan x + arctan y = π + arctan (x+y/1-xy) , x,y>0 and xy >1
If x > 0, y > 0 such that xy > 1, then x+y1−xyx+y1−xy is positive and therefore, x+y1−xyx+y1−xy is positive angle between 0° and 90°.
Similarly, if x < 0, y < 0 such that xy > 1, then x+y1−xyx+y1−xy is positive and therefore, tan−1−1 (x+y1−xyx+y1−xy) is a negative angle while tan−1−1 x + tan−1−1 y is a positive angle while tan−1−1 x + tan−1−1 y is a non-negative angle. Therefore, tan−1−1 x + tan−1−1 y = π + tan−1−1 (x+y1−xyx+y1−xy), if x > 0, y > 0 and xy > 1 and
arctan(x) + arctan(y) = arctan(x+y1−xyx+y1−xy) - π, if x < 0, y < 0 and xy > 1.
hope this is helpful
hardiknegi99:
plz.............. don't copy from Google................. If u really know then explain it
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