Pre Calculus Trig Proof

Question

I dont get the approach on how to prove this statement is true

(tan x)/(1-cot x) + (cot x)/(1-tan x) = 1 + sec x(csc x)

Answer 1

tanx/(1-cotx) + cotx/(1-tanx) = 1+ secx(cscx)

if you don't know what to do, one method is convert everything to sines and cosines, eliminate fractions and simplify. Although there may be easier ways to do this problem.

(sinx/cosx)/(1-cosx/sinx) + (cosx/sinx)/(1-sinx/cosx) = 1 +(1/cosx)(1/sinx)= 1+1/cosxsinx

(sinx/cosx)/((sinx-cosx)/sinx) + (cosx/sinx)/((cosx-sinx)/cosx) = (cosxsinx + 1)/cosxsinx

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

multiply by sinxcosx(sinx-cosx), the least common denominator to eliminate fractions

sin^3x - cos^3x = (cosxsinx + 1)(sinx - cosx)

factor sin^3x - cos^3x (general formula is a^3 - b^3 = (a-b)(a^2+ab+b^2) let a=sinx, b=cosx)

(sinx-cosx)(sin^2 +sinxcosx + cos^2x) = (cosxsinx+1)(sinx-cosx)

cancel sinx-cosx from both sides (or view it as dividing both side by the factor sinx-cosx, leaving

sin^2 + sinxcosx + cos^2x = cosxsinx +1

cancel sinxcosx from both sides, or view it as subtracting sinxcosx from both sides, leaving

sin^2x + cos^2x = 1 which is a form of the Pythagorean Theorem and identity, always true regardless of the value of angle x

1= 1

QED

Answer 2

Multiply the left side of the equation by the common denominator:

(1 - cot x)(1 - tan x), then combine and simplify.