(aSinA-bCosA)/(aSinA+bCosA)

Question

Given that, CotA = b/a

Answer 1

(a sin A - b cos A) / (a sin A + b cos A)

 

Divide both the numerator and denominator by sin A:

 

[ (a sin A - b cos A) / sin A ] / [ (a sin A + b cos A) / sin A ]

 

Simplify the numerator and denominator:

(a sin A / sin A - b cos A / sin A) / (a sin A / sin A + b cos A / sin A) = 

(a - b cos A / sin A) / (a + b cos A / sin A) =

(a - b cot A) / (a + b cot A)

 

Substitute b/a for cot A:

(a - b * b/a) / (a + b * b/a) =

(a - b²/a) / (a + b²/a)

 

Divide numerator and denominator by a:

(1 - b²/a²) / (1 + b²/a²)

 

Substitute cot A for b/a  (that is, substitute cot² A for b²/a²

(1 - cot² A) / (1 + cot² A)

 

Remember that 1 + cot²A = csc² A = 1 / sin² A

 

So our expression is equal to:

(1 - cot² A) / (1 / sin² A) =

sin² A (1 - cot² A) =

sin² A  (1 - cos² A / sin² A) =

sin² A - cos² A =

- (cos² A - sin² A) =

- cos 2A

 

I might have missed something along the way, but that's what I get.

 

 

 

 

 

 

Answer 2

(aSinA-bCosA)/(aSinA+bCosA)

 

let us divide all terms by aSinA

 

= (1- bCosA/aSinA)/(1+ bCosA/aSinA)

 

= (1 - (b/a)CotA)/(1 + (b/a)CotA)

 

= (1 - (b/a)(b/a))/(1 + (b/a)(b/a))

 

= (1 - (b/a)²)/(1 + (b/a)²)

 

= (a² - b²)/(a² + b²)