Thursday

August 28, 2014

August 28, 2014

Posted by **abdo** on Wednesday, April 18, 2007 at 2:37pm.

(2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx)

and thanks ...........

check your typing.

I tried 30º, the two sides are not equal, they differ by 1

oh , thank you Mr Reiny I'll tell my teacher this Question is Wrong .

not necessarily.

look at your first term

what does tan^x mean?

I want you to check what you typed here with your question.

Mr Reiny ,

I tried 30º, the two sides are equal

L.S=3.732

R.S=3.732

Yes, if the term at the bottom is tan^2 x,

you typed tan^x, and I read it as tan x

I got the proof, give me a bit of time to type it

I'm sorry Mr Reiny you are right it is tan^2x .

LS=

2tanx/(1-tan^2x) + 1/(2cos^2x -1)

= 2sinx/cosx [1/(1-sin^2x/cos^2x)] + 1/(2cos^2x -(sin^2x + cos^2x))

=2sinx/cosx [cos2x/(cos^2x - sin^2x)] + 1/(cos^2x - sin^2x) reduce to get same denominator

=2sinxcosx/(cos^2x-sin^2x) + 1/(cos^2x-sin^2x)

=(2sinxcosx + sin^2x + cos^2x)/(cos^2x-sin^2x)

= (cosx+sinx)(cosx+sinx)/[cosx+sinx)(cosx-sinx)]

=(cosx+sinx)/(cosx - sinx)

= Right Side!!!!!

thank you Mr Reiny

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