If tanθ=ab, then the value of E =asinθ−bcosθasinθ+bcosθ is−
(a2 – b2)/(a2 + b2)
(a-b)/(a+b)
(a+b)/(a-b)
(a-b)/(a2 + b2)
(a2 + b2)/(a2 - b2)
Dividing both the numerator and the denominator of E by cos θ, we have :
E=atanθ−batanθ+b=a(ab)−ba(ab)+b=a2−b2a2+b2
If tanθ=ab, show that asinθ−bcosθasinθ+bcosθ=a2−b2a2+b2