Question

Another proving of identites:
tanx/(tan^2x-1) = 1/(tanx-cotx) ?
How?

Answer 1

more algebra

Answer 2

sir thanks in advance. how it can be solved sir?

Answer 3

hold on, let me do some damned algebra

Answer 4

hahah. sorry sir. ok sir.

Answer 5

ok lets replace,as usual, \(\cos(x)\) by \(a\) and \(\sin(x)\) by \(b\)

in this example there is literally no trig at all! it is nothing but straight up algebra
no trig identities after we write everything in terms of sine and cosine

Answer 6

ok sir.

Answer 7

making that replacement, i get on the left
\[\frac{\frac{b}{a}}{\left(\frac{b}{a}\right)^2-1}\] and stop calling me sir

Answer 8

wait... no trig identities? wow.

Answer 9

on the right, i get
\[\frac{1}{\frac{b}{a}-\frac{a}{b}}\]

Answer 10

ok i guess i don't mean actually no trig identities, but the only ones i used are rewriting everything in terms of sine and cosine
that is all

Answer 11

ok ok.

Answer 12

using basic algebra you can show rather easily that
\[\large \frac{\frac{b}{a}}{\left(\frac{b}{a}\right)^2-1}=\frac{1}{\frac{b}{a}-\frac{a}{b}}\] as they are both equal to
\[\frac{ab}{b^2-a^2}\]

Answer 13

for the right hand side, multiply top and bottom by \(ab\) and you get it in one step

Answer 14

hmmm. (looking a the post)

Answer 15

for the left hand side, multiply top and bottom by \(a^2\) and you get the same thing

Answer 16

umm. can't follow, why should I multiply the right side by ab and left side by a^2? where did it came from?

Answer 17

ok are you good to this step \[\large \frac{\frac{b}{a}}{\left(\frac{b}{a}\right)^2-1}=\frac{1}{\frac{b}{a}-\frac{a}{b}}\]

Answer 18

yes

Answer 19

replaced all cosines by \(a\) and all sines by \(b\)
ok good

Answer 20

both sides are an ugly mess, i.e. they are both compound fractions, or complex fractions, or whatever they are called
too many bars in other words

Answer 21

yes. true. I honestly don't know what to do next after that.

Answer 22

so imagine you were in an algebra class and came upon
\[\frac{1}{\frac{b}{a}-\frac{a}{b}}\] what would you do?

Answer 23

reciprocal then cross multiply?

Answer 24

clear the fractions by multiplying top and bottom by \(ab\)

Answer 25

i think what you mean really is add the fraction in the denominator and then take the reciprocal
that would work too, but it is the donkey way
cowboy way is to do this
\[\frac{1}{\frac{b}{a}-\frac{a}{b}}\times \frac{ab}{ab}=\frac{ab}{b^2-a^2}\]

Answer 26

mmm. will the answer becomes ab/(b^2-a^2)?

Answer 27

yes

Answer 28

now how about this ugly beast
\[\large \frac{\frac{b}{a}}{\left(\frac{b}{a}\right)^2-1}\]

Answer 29

ooohhh. now I get it.

Answer 30

or if you prefer
\[\large \frac{\frac{b}{a}}{\frac{b^2}{a^2}-1}\]

Answer 31

hahah ugly beast. yeah, why did you come up with a^2?

Answer 32

now you can see it right? why \(a^2\) will clear the fraction?

Answer 33

hmmm.

Answer 34

it is the common denominator of \(a\) and \(a^2\)

Answer 35

\[\large \frac{\frac{b}{a}}{\frac{b^2}{a^2}-1}\times \frac{a^2}{a^2}=...\] no more complex fraction

Answer 36

ah ok-ok. so that's how it is done. so the answer is ab/(b^2-a^2) ?

Answer 37

thanks sir.

Answer 38

yup
as you can see, it was all algebra, no trig
at least once you write all in terms of sine and cosine

Answer 39

imagine trying to do that algebra writing sine and cosine every time
would make the confusing algebra almost impossible right?

Answer 40

thanks for the technique sir. I will use it and solve the 54 identities in the book.
:)

hahahah. yeah.