Question

Tan θ =(a/b)
then, prove that {(a sinθ -b cos θ)/(a sinθ +b cos θ )} = {(a^2-b^2)/(a^2+b^2)}

Answer 1

Tan0 = sing0/cos0 =a/b
so sing0=(a cos0)/b
put sing0 value in your given equation 

(a*a cos0/b - b cos0)/ (a*a cos0/b + b cos0)
next step
=(a^2 cos0 -b^2 cos0 )/ (a^2 cos0 + b^2 cos0)
=cos0(a^2 - b^2) / cos0(a^2 + b^2)
here cos0 is divided by cos0 so answer 
= (a^2 - b^2) / (a^2 + b^2)

Answer 2

Here's your ans...

Hope it helps.....