Question

Prove the Trig. Identitiy:
(1+tanx)/(1+cotx)=tanx

Answer 1

1 + sin x
----
cos(x)
-------------------------
1+ cos(x)
------
sin(x)

cos(x)+sin(x) sin(x)
------------ *
cos(x) -------------
sin(x)+ cos(x)


sin(x)
-----= tan(x)
cos(x)

Answer 2

How did you get from step one to step two?

Answer 3

found the common denominator

Answer 4

division is same as multiplication by reciprocal

cos(x)+sin(x)
------------
cos(x)

---------------- --------------
sin(x)+ cos(x)
--------------
sin(x)


so we flip the denominator


cos(x)+sin(x) Sin(x)
------------ * --------------
cos(x) cos(x)+sin(x)

Answer 5

Still confused?

Answer 6

Yes, slighltly....?

Answer 7

Slightly*

Answer 8

what's seems troubling?

Answer 9

Where did the 1 in the denominator go from step 1 to step 2?

Answer 10

common denominator
1 + sin x
----
cos(x)

cos(x) + sin(x)
-------------
cos(x)

Answer 11

can i butt in? you know
\[\tan(x)=\frac{\sin(x)}{\cos(x)}\] and
\[\cot(x)=\frac{\cos(x)}{\sin(x)}\] now replace cosine by "a" sine by "b" and show that
\[\frac{1+\frac{b}{a}}{1+\frac{a}{b}}=\frac{b}{a}\]

Answer 12

cos(x)+sin(x) sin(x)
------------ *
cos(x) -------------
sin(x)+ cos(x)

This is just throwing me all off. Sorry. :\

Answer 13

it is pure algebra and has nothing at all to do with sine and cosine

Answer 14

sin(x)+cos(x)= cos(x)+sin(x)

thus they cancel

Answer 15

so ignore them. it is much easier to write
\[\frac{1+\frac{b}{a}}{1+\frac{a}{b}}=\frac{b}{a}\] then the equivalent stuff with a bunch of trig functions. this is my best suggestion for doing these problems. do the algebra with variables and then put the trig functions back in

Answer 16

i wouldn't even bother with adding this stuff up. i would multiply top and bottom by
\[ab\] and clear the fractions.

Answer 17

\[\frac{1+\frac{b}{a}}{1+\frac{a}{b}}\times \frac{ab}{ab}\]
\[\frac{ab+b^2}{ab+a^2}\]
\[\frac{b(b+a)}{a(b+a)}=\frac{b}{a}\]

Answer 18

done in three steps.

Answer 19

Oh my goodness, @satellite73, thank you so so so very much. (:

Answer 20

i hope it is clear that this particular problem has nothing whatsoever to do with trigonometry. it would work with numbers, variables, two other functions, anything

Answer 21

Yes, I was trying way too hard to apply a trig ideas. Inserting a's and b's helped a lot. (: