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

Question

Please Prove it

Answer 1

RHS:

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

LHS:

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

Answer 2

Multiply it out. You get: secx - sinxsecx + tanx -sinxtanx = a million/cosx -sinx/cosx + tanx - sin^2 (x) /cosx = a million/cosx - tanx + tanx - sin^2(x) /cosx = (a million - sin^2(x)) / cosx = cos^2(x) /cosx = cosx QED

Answer 3

LHS=[(sinx+cosx)/sinxcosx]/(sinx+cosx)
=1/sinxcosx
=secxcscx

RHS
=sec^2x/tanx
=sec^2x*cosx/sinx
=secxcscx
=LHS
hence proved

Answer 4

[cscx +secx]/[sinx +cosx] = [1+tan^2x]/tanx
[1/cosx +1/sinx]/[sinx+cosx] = [1+tan^2x]/tanx
[(sinx +cosx)/sinxcosx]/[sinx +cosx]=[1+tan^2x]/tanx
1/(sinxcosx) = (1+sin^2x/cos^2x)/(sinx/cosx)
1/(sinxcosx) = [(cos^2x + sin^2x)/cos^2x]/(sinx/cosx)
1/(sinxcosx) = [1/(cos^2x)](cosx/sinx)
sinx/cos^3x = sinx/cos^3x
1=1