Question

If tan theta =a/b prove that, a sin theta - b cos theta/ a sin theta +b cos theta = a 2 - b 2 / a 2 + b 2

Answer 1

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: a = b = θ/2 

cos[(θ/2) + (θ/2)] = cos(θ/2).cos(θ/2) - sin(θ/2).sin(θ/2) 

cos(θ) = cos²(θ/2) - sin²(θ/2) → recall: cos²(θ/2) + sin²(θ/2) = 1 → sin²(θ/2) = 1 - cos²(θ/2) 

cos(θ) = cos²(θ/2) - [1 - cos²(θ/2)] 

cos(θ) = cos²(θ/2) - 1 + cos²(θ/2) 

cos(θ) = 2.cos²(θ/2) - 1 

2.cos²(θ/2) = cos(θ) + 1 

cos²(θ/2) = [cos(θ) + 1]/2 ← memorize this result 



= [tan(θ) + sin(θ)] / [2.tan(θ)] → recall: tan(θ) = sin(θ)/cos(θ) 

= [ {sin(θ)/cos(θ)} + sin(θ)] / [2.{sin(θ)/cos(θ)}] 

= [ {sin(θ)/cos(θ)} + {sin(θ).cos(θ)/cos(θ)} ] / [2.{sin(θ)/cos(θ)}] 

= [ {sin(θ) + sin(θ).cos(θ)}/cos(θ) ] / [2.{sin(θ)/cos(θ)}] → you can simplify by cos(θ) 

= [sin(θ) + sin(θ).cos(θ)] / [2.sin(θ)] → you factorize sin(θ) 

= sin(θ).[1 + cos(θ)] / [2.sin(θ)] → you can simplify by sin(θ) 

= [1 + cos(θ)]/2 → recall the memorized result 

= cos²(θ/2) ← be careful, you made a mistake: it's cos²(θ/2) instead of cos²(θ)la console · 2 months ago