Help me prove this trig identity!!!!?

Question

first one is (cosx/(1-tanx)) + (sinx/(1-cotx)) = sinx + cosx

second identity is (xsinA + ycosA)^2 + (xcosA - ysinA)^2 = x^2 + y^2

All those math geniuses out there, please solve these two I have no idea what to do

Answer 1

#1:
cosx/(1-tanx)) + (sinx/(1-cotx))
= cosx/(1-sinx/cosx) + sinx/(1-cosx/sinx)
= cos²x/(cosx - sinx) + sin²x/(sinx-cosx)
= cos²x/(cosx - sinx) - sin²x/(cosx - sinx)
= [cos²x - sin²x]/(cosx - sinx)
= (cosx + sinx)(cosx - sinx)/(cosx - sinx)
= (cosx + sinx)


#2: Show (xsinA + ycosA)² + (xcosA - ysinA)² = x² + y²
(xsinA + ycosA)² + (xcosA - ysinA)² =
(x²sin²A +2xysinAcosA + y²cos²A) + (x²cos²A -2xycosAsinA + y²sin²A) =
x²sin²A + y²cos²A + x²cos²A + y²sin²A =
x²(sin²A + cos²A) + y²(cos²A + sin²A) =
x² + y²

Answer 2

#1 LHS: change to sines/cosines and get rid of fractions
cosx / (1 - sinx/cosx) ===> multiply top and bottom to clear fractions
=cos^2(x) / (cosx - sinx)
similar
sinx/(1-cosx/sinx)
=sin^2(x) / (sinx -cosx)
add them, we need LCD
cos^2(x) / (cosx - sinx) - sin^2(x) / (cosx - sinx)
[cos^2(x) - sin^2(x)] / (cosx -sinx)
factor numerator diff of perfect squares
(cosx - sinx) ( cosx + sinx) / (cosx - sinx) ===> cancel
cosx + sinx

:)


#2
square each term(FOIL)
x^2sin^2 (A) +2xycosAsinA + y^2cos^2(A) + x^2cos^2 (A) - 2xycosAsinA + y^2sin^2(A)
= x^2sin^2(A) + x^2cos^2(A) +y^2cos^2(A) + y^2sin^2(A)
= x^2 [ sin^2(A) + cos^2(A) ] + y^2 [ cos^2(A) + sin^2(A) ]
because Pyth ID ===> cos^2 + sin^2 =1
x^2 + y^2

:)

Answer 3

Question 1
cosx / (1 - sinx / cosx) + sinx / (1 - cosx / sinx)
= cos²x / (cosx - sinx) + sin²x / (sinx - cosx)
= cos²x / (cosx - sinx) - sin²x / (cosx - sinx)
= (cos²x - sin²x) / (cosx - sinx)
= (cosx - sinx).(cosx + sinx) / (cosx - sinx)
= cosx + sinx
Question 2
Squaring the brackets gives:-
x².sin²A + 2.x.y.sinA.cosA + y².cos²A
x².cos²A - 2.x.y.sinA.cosA + y².sin²A
Adding gives:-
x².(sin²A + cos²A) + y².(cos²A + sin²A)
= x² + y²

Answer 4

multiply 1st fraction by cos/cos [I'll leave out all the x's: I'm lazy] and the 2nd by sin/sin, getting

cos²/(cos - sin) + sin²/(sin - cos) =
sin²/(sin - cos) - cos²/(sin - cos) =
(sin + cos)(sin - cos)/(sin - cos)=
(sin x + cos x)

2nd one just needs multiplying out the square binomials. I'll put 2nd under 1st:

x²sin² + 2xy•sin•cos + y²cos²
x²cos² - 2xy•sin•cos + y²sin²
------------------------------------
x²(sin² + cos²) + y²(cos² + sin²)
x² + y²