[ csc(x)+sec(x) ] / [sin(x)+cos(x)]= [1+ tan^(2) (x)] / tan(x)?

Question

Please prove it, showing all your work

Answer 1

What's the point of asking the same question twice?

If the issue is that you're only allowed to work on one side of the equation or the other, then you can pick one side, follow down to the bottom, and do everything in reverse to match the other side.

RHS:

1 + tan²x / tanx = [1 + (sec²x - 1)] / tanx = sec²x / tanx
= 1/cos²x / (sinx / cosx) = 1/cos²x * cosx/sinx
= 1/(sinx cosx) = cscx secx

LHS:

[cscx + secx]/[sinx + cosx] = [1/sinx + 1/cosx]/[sinx + cosx]
= [(cosx + sinx)/(sinx cosx)] * 1/[sinx + cosx]
= 1/(sinx cosx) = cscx secx

Answer 2

It Reallyyyy easy....i am not gonna solve it...i'll tell u how to do it...it's really easy! These are some trigonomertic formulas: sin, cos, tg, ctg, sec, csc: sinus, cosinus, tangentus, cotangentus, secant, cosecant. They have some values given for 30, 45, 60 degrees. If you know those values, no problemo. Take all the values for 30 degrees and replace (x) with 30 degrees. you will get [csc(30)+sec(30)]/[sin(30)+cos(x)]=[1+tan^(2)(30)]/tan(30). You can find these values anywhere...Good Luck

Answer 3

rhs = [sec (^2) x]/tanx = [(cosx) ^-2] * (cosx / sinx) =
(sinxcosx) ^-1 = 1/(sinxcosx)

lhs = [(sinx)^-1 + (cosx)^-1](sinx +cosx) = 1/(sinxcosx) (on taking lcm & cancelling Sinx + cosx term )

Using
1 + tan square x = sec square x
secx = 1/cosx
cosecx = 1/sinx

Answer 4

cscx+secx=1/sinx+1/cosx
taking LCD sinxcosx
cscx+secx=sinx+cosx/sinxcosx
cscx+secx/(sinx+cosx)
=[sinx+cosx/sinxcosx]/(sinx+cosx
=1/sinxcosx
=(1/sinx)*(1/cosx)
=cscxsecx
now if we are able to reduce (1+tan^2x)/tanx also to
cscxsecx we would have proved
1+tan^2x=sec^2x (identity)
=1/cos^2x
tanx=sinx/cosx
so (1+tan^2x)/tanx
=(1/cos^2x)/sinx/cosx)
flipping the denominator
=(1/cos^2x)(cosx/sinx)
=1/sinxcosx
=(1/cosx)(1/sinx)
=cscxsecx

Answer 5

What the hell is this?