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Sourman
Jun 19, 2012, 01:26 PM
can anyone verify this

1+tanX/1+cotX = 1-tanX/cot-1

please include all steps

ebaines
Jun 19, 2012, 02:05 PM
1+tanX/1+cotX = 1-tanX/cot-1

Please use parentheses! What you wrote is equivalent to:

1+tanx +cotx = -tanx/cotx

which is not accurate. What I think you meant is this:
(1+tanx)/(1+ cot x) = (1-tanx)/(cotx - 1)

You could try cross-multiplying and then converting the cot(x) and tan(x) functions to their sin(x) and cos(x) equivalents.

Sourman
Jun 19, 2012, 02:57 PM
Please use patentheses! What you wrote is equivalent to:

1+tanx +cotx = -tanx/cotx

which is not accurate. What I think you meant is this:
(1+tanx)/(1+ cot x) = (1-tanx)/(cotx - 1)

You could try cross-multiplying and then converting the cot(x) and tan(x) functions to their sin(x) and cos(x) equivalents.

Ya sorry this is the equation I meant, however if I do the method you proposed would I not be breaking the rules of proving trig Identities, by cross multiplying am I not then assuming that the equation is already equal? I thought when doing trig identities you must work on each side seperatly in order to prove their equality.

ebaines
Jun 19, 2012, 03:05 PM
I hadn't heard of that rule, but no matter. Try this - convert the cot(x) and tan(x) functions to their sine and cosine equivalents. Then simplify, first by multiplying through both numerator and denominator by sin(x)cos(x) and then gathering and eliminating like terms. Comes right out.

Sourman
Jun 19, 2012, 03:40 PM
I hadn't heard of that rule, but no matter. Try this - convert the cot(x) and tan(x) functions to their sine and cosine equivalents. Then simplify, first by multiplying through both numerator and denominator by sin(x)cos(x) and then gathering and eliminating like terms. Comes right out.

K I was finally bale to get it, thanks for the help!