How do I solve this trig identity: cos2x/(1+sinx^2)=tan(pie/4 -x)

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How do I solve this trig identity: cos2x/(1+sinx^2)=tan(pie/4 -x)

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i think ur question is cos2x/(1+sin2x) now cos2x =cos^2x -sin^2x=(cosx+sinx)(cosx-sinx) and 1+sin2x = cos^2x + sin^2x +2sinxcosx =(cosx+sinx)^2 thus cos2x/(1+sinx)^2= (cosx+sinx)(cosx-sinx)/(cosx+sinx)^2 =(cosx- sinx)/((cosx+sinx) divide both num and denom by cosx and then we have =(1-sinx/cosx)/(1+sinx/cosx) =(1-tanx)/(1+tanx) = (tan(pi/4) -tanx)/(1+tan(pi/4)tanx)=tan(pie/4 -x)
for the second last step.. where did the square roots go? isnt it suppose to be: (1-sinx/cosx)(1-sinx/cos)/(1+sinx/cos)(1+sinx/cosx)
actually the factors (cosx+sinx) are there both in num and denom hence they cancel each other

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Other answers:

but its (cosx-sinx) in the num :s
(cosx+sinx)(cosx-sinx) / (cosx+sinx)^2 =(cosx-sinx) / (cosx+sinx)
i hope its clear now
why 13 u hv included in ur id ? if u don't mind
if its (cosx+sinx)^2, it has to equal (cosx+sinx)(cosx+sinx)
idk random.
(cosx+sinx)^2= (cosx+sinx)(cosx+sinx) (correct ) that is why one of the (cosx+sinx)
cancel each other
how does cos^2x -sin^2x=(cosx+sinx)(cosx-sinx)
USE ..A*A -B*B =(A)^2 -(B)^2 = (A+B)(A-B)...
okay, then why isnt :(cosx+sinx)^2 --> (cosx-sinx)(cosx+sinx)
bcoz a^2 =a*a
so (a+b)^2 =(a+b)(a+b)
but if u multiply (a+b)(a-b) u get (a^2 -b^2)
so (a^2-b^2) does not equal (a-b)^2 ?
sahi hai..
oh i get it..
=(1-tanx)/(1+tanx) = (tan(pi/4) -tanx)/(1+tan(pi/4)tanx)=tan(pie/4 -x) can u explain that? .. where did pi/4 come from ?
Tan (pi/4) = 1.
is that a trig identity?
No. If you do that on a calculator you get that.
Tan(pi/4) is always 1. Look on a unit circle.
Okay, thanks!!
Welcome :) .

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