a trig idenity question?

Question

if a / sinA = b / cosA, show that

sinAcosA = ab / (a^2+b^2)

another one:

if (a+b)/cosecx = (a-b)/cotx shown that

cosecx cotx = (a^2 - b^2) / 4ab

Answer 1

if (a+b)/cscx = (a-b)/cotx show that EQ 1

cscxcotx = (a^2 - b^2) / 4ab

From eq 1:

cotxcscx = [ (a - b)cscx/(a+b)]*cscx
= (a - b)csc^2(x)/(a+b) EQ 2

If we draw a right triangle in which hypotenuse = a + b and adjacent = a - b, then we use the Pythagorean Theorem to find the opposite = sqrt[ (a + b)^2 - (a - b)^2 ] = 2sqrt(ab)

So, taking in mind that cscx = 1/sinx = hyp/opp
then cscx = ( a + b) / 2sqrt(ab) therefore:

csc^2(x) = (a + b)^2 / 4ab EQ 3

Plugging in EQ 3 at EQ2:

cotxcscx = (a - b)csc^2(x)/(a+b)
= (a - b)(a + b)^2 / 4ab(a+b)
= (a - b)(a + b) / 4ab
= (a^2 - b^2) / 4ab

Answer 2

a/sinA = b/cosA

multiply both sides by (sinA)(sinA)(cosA)/a,
sin A cosA = b (sinA)^2 /a

(sin A) = a/sqrt (a^2 + b^2)
--> sin of an angle is equal to opposite side (a) divided by its hypotenus (sqrt (a^2 + b^2)..

(sin A)^2 = a^2 / (a^2 + b^2)

substitute:
sin A cos A = b (a^2 / (a^2 + b^2) / a
sin A cos A = ab / (a^2 + b^2)