Question

how to prove
tan^(-1) x +tan(^-1) y= tan^(-1) (x+y)/(1-xy)

Answer 1

\(\color{#000000 }{ \displaystyle \tan^{-1}(x)+\tan^{-1}(y)=\frac{\tan^{-1}(x+y) }{1-xy}}\)

like this?

Answer 2

the answer is \[\tan ^{-1} \frac{ x+y }{ 1-xy }\]

Answer 3

\(\color{#000000 }{ \displaystyle \tan^{-1}(x)+\tan^{-1}(y)=\tan^{-1}\left(\frac{x+y }{1-xy}\right)}\)

Like that you mean?

Answer 4

yup

Answer 5

I think you mistranslated @mathmale .

Answer 6

try taking tan( ) of both sides @eunna

Answer 7

then?

Answer 8

Do you know sum rule for tan?

Answer 9

no

Answer 10

What identities are you allowed to use then?

Answer 11

ohh the one that tan(a+b)

Answer 12

yes

Answer 13

a+b is read as the sum of a and b

Answer 14

after apply the sum rule, how can i make it inverse?

Answer 15

do you know what tan(arctan(u)) equals?

Answer 16

equal to u

Answer 17

right

Answer 18

so i just need to multiply with tan^-1

Answer 19

wont it eliminate the tan

Answer 20

multiply?

Answer 21

Have you taken tan( ) of both sides and applied sum rule yet for tan? and then use that tan(arctan(u))=u?

Answer 22

If you are having trouble understanding this way, you could try doing the following:
Let arctan(x)=u and arctan(y)=v
and so tan(u)=x and tan(v)=y...
Plug tan(u)=x and tan(v)=y into the right hand side...
You will still have to use sum rule for tan...

Answer 23

for example you know
\[\frac{\tan(u)+\tan(v)}{1-\tan(u)\tan(v)}=?\]

Answer 24

i got it, thank you so much

Answer 25

np