Trig question?

Question

I need to be able to verify this equation.
You can only modify one side though.

( tan(x) + sec(x) ) / ( cos(x) + cot(x) ) = sec(x) tan(x)

I have been working on this, and i just get a big mess.

I need the step by step

Thanks in advance

Answer 1

(tan+ sec ) / (cos + cot ) =
[( sin + 1 ) / cos ] / [( sin cos + cos)/sin ] =
[sin / cos ] [ sin + 1 ] / [ (sin + 1 ) cos ] =
[ sin / cos ] / cos =
sec tan

Answer 2

You can modify both sides, but separately. It's not an equation until you prove both expressions are equal. Don't feel bad about the big mess. I went through 3 big messes before I found my way through.

On the right, using basics, [and forgive me for leaving out x, too many years teaching made me lazy, and as long as we don't have half or double angles....],

sec • tan = 1/cos • sin/cos = sin/cos²

on the left,

[tan + sec] / [cos + cot] =
[sin/cos + 1/cos] / [cos + cos/sin] =
[(sin + 1)/cos] / [cos(1 + 1/sin)] =
[sin + 1] / [cos² ( sin + 1)/sin] =
1 / [ cos² / sin] =
sin / cos²

and that's the identity

Answer 3

(tan+sec)= (sin/cos + 1/cos) = (sin+1) / cos
(cos+cot) = (cos + cos/sin) = (cossin + cos) / sin
so left side = (sin(sin+1)) / (cos(cossin+cos)
(by dividing (sin+1)/cos by (cossin + cos) / sin)

= (sin(sin+1))/cos(cos(sin+1)))
so we can remove both sin+1, leaving
sin / cos^2 = (sin/cos)(1/cos) = tan(x)sec(x)

the trick is to convert everything to sin and cos and try to work with them.

Answer 4

rewrite using sin and cos.

((sinx/cosx)+(1/cosx))/(cosx+(cosx/sinx))

=((sinxcosx+cosx)/cos^2(x))*(sinx/(cosxsinx+cosx))

Cancel appropriately:

sinx/cos^2(x)=(sinx/cosx)*(1/cosx)=tanx*secx

Answer 5

(sinx/cosx+1/sinx)/(cosx+cosx/sinx)=1/cosx
[[(sinx)^2+cosx)/cosxsinx]/
[(cosxsinx+cosx)/sinx]]=1/cosx
(sinx)^2+cosx=cosxsinx+cosx
sinx(sinx-cosx)=0
sinx=0<-->x=npi
sinx-cosx=0
sinx=cosx
tanx=0
x=npi
nEZ