If a/sinA=b/cosA, show that sinA cosA = ab/a^2+b^2

Any help is much appreciated! I've spent the last 2 hours on this question..

Given : a / sin A = b / cos A = k, say ..... (1)

∴ a = k. sin A, b = k. cos A ................... (2)

∴ a² + b² = k² ........................................… (3)

∴ from (2),

sin A cos A = ( a/k )( b/k )

. . . . . . . . . .= ( ab ) / k²

. . . . . . . . . .= ( ab ) / ( a² + b² ) ........ from (3)

Hence, the result.
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if a/sinA = b/cosB then

sinA/cosB = a/b by simple rearrangement

tanA = a/b as sinA/cosA = tanA

Draw a right angled triangle with A as one of the angles.

tan theta = opposite/adjacent so label the opposite side of the triangle with a and the adjacent side with b.

use Pythagoras to get a hypotenuse of sqrt(a^2 +b^2)

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

when you multiply fractions you multiply the numerators to get the new numerator a *b = ab
and multiply the denominators (sqrt(a^2 +b^2) )^2 = a^2 +b^2

thus cosA sin A = ab / a^2 + b^2

squaring the given eq
a^2/sin^2a =b^2/cos ^2 a [cos^2 a = 1- sin^2 a]
a^2/sin^2 a = b^2 / 1-sin^2 a
solving we get
sin^2 a = a^2 / (a^2 + b^2)....... (i)
similarly as sin^2 a = 1- cos^2 a therefore
cos^2 a= b^2/ (a^2 + b^2)..........(ii}
multiplying eq (i) and (ii)
cos^2 a x sin^2 a = a^2 * b^2/(a^2 + b^2)^2
cos a * sin a = ab/(a^2 + b^2) [square rooting both lhs and rhs of above eq.]

hope u'll appriciate our answer

a/sinA=b/cosA multiply by a/cos A both sides
a^2/(sin A cos A) = ab / cos^2 A
multiply first equality by b/sin A
ab/sin^2 A = b^2/(sinA cos A)
Now
a^2/(sin A cos A) = ab / cos^2 A
b^2/(sinA cos A) = ab/sin^2 A
Or
(sin A cos A) / a^2 = cos^2 A /ab
(sinA cos A) /b^2 =sin^2 A/ab
========================= add them
(sin AcosA)(1/a^2+1/b^2) = 1/ab(sin^2 A+sin^2 A) but sin^2 A+sin^2 A = 1
(sin AcosA)(1/a^2+1/b^2) = 1/ab
sin A cosA = 1/ab * a^2b^2/(a^2+b^2) = ab/(a^2+b^2)
QED

i have worked out in pen and paper format here
https://docs.google.com/leaf?id=1ed8sVkVUguI8K6h46e4CXx4RqfiU54JH2xXfaI-tIx8&hl=en_US&authkey=COa-uI0D