why is this method of solving $\tan^{-1}(\frac{x}{y})-\tan^{-1}(\frac{x-y}{x+y})$ wrong?

Asked Oct 06 '22 at 12:26

Active Oct 06 '22 at 12:32

Viewed 42 times

$$\tan^{-1}\left(\frac{x}{y}\right)-\tan^{-1}\left(\frac{x-y}{x+y}\right)$$ let this be equal to $A$ so $\tan^{-1}(\frac{x}{y})-\tan^{-1}(\frac{x-y}{x+y})=A$ thus $$\frac{x(x+y)-y(x-y)}{y(x+y)+x(x-y)}=\tan(A)$$ which means that the original function is dependent on both $x$ and $y$,

however, my book says it's independent of both.

I'd really appreciate some help.

edited Oct 06 '22 at 12:32

asked Oct 06 '22 at 12:26

math and physics forever

1,200

How did you go the the last equation from $A$ to $\tan(A)$? – Gary Oct 06 '22 at 12:28
I took tan on both sides – math and physics forever Oct 06 '22 at 12:28
Numerator should have x(x+y) and denominator should have y(x+y). – insipidintegrator Oct 06 '22 at 12:29
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Use this on the left-hand side. – Gary Oct 06 '22 at 12:30
that's what I used – math and physics forever Oct 06 '22 at 12:31
@insipidintegrator, I thought I had edited that, I''l change it now – math and physics forever Oct 06 '22 at 12:31
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Yeah, then after simplifying you get tan A =1. – insipidintegrator Oct 06 '22 at 12:32
oh, in that case I must have made a calculation error, thanks! – math and physics forever Oct 06 '22 at 12:36
https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan – lab bhattacharjee Oct 06 '22 at 12:38

why is this method of solving $\tan^{-1}(\frac{x}{y})-\tan^{-1}(\frac{x-y}{x+y})$ wrong?

0 Answers0