Answer:
<h2>
tan (A−B)=tan(A)−tan(B)/1+tan(A)tan(B)</h2>
tan(A−B)=tan(A)−tan(B)/1+tan(A)tan(B)
1+tan(A)tan(B) To prove this, we should know that:
1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)
1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)AND
1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)ANDcos(A−B)=cos(A)cos(B)+sin(A)sin(B)
1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)ANDcos(A−B)=cos(A)cos(B)+sin(A)sin(B)We start with:
1+tan(A)tan(B) To prove this, we should know that:sin(A−B)=sin(A)cos(B)−cos(A)sin(B)ANDcos(A−B)=cos(A)cos(B)+sin(A)sin(B)We start with:tan(A−B)=sin(A−B)/cos(A−B)
cos(A−B)tan(A−B)=sin(A)/cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)
cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)
cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,
- cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0
- cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0
- cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0
- cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0cos(A)≠0 and cos(B)≠0
cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0cos(A)≠0 and cos(B)≠0The question on this page is asking for specific formula, which comes from Case 4, so I will only solve Case 4 here.
cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0cos(A)≠0 and cos(B)≠0The question on this page is asking for specific formula, which comes from Case 4, so I will only solve Case 4 here.Dividing numerator and denominator in (1) by cos(A)cos(B) we get:
cos(B)−cos(A)sin(B)cos(A)cos(B)+sin(A)sin(B)(1)We would want to divide both numerator and denominator by cos(A)cos(B)but there are 4 cases,cos(A)=0 and cos(B)≠0cos(A)≠0 and cos(B)=0cos(A)=0 and cos(B)=0cos(A)≠0 and cos(B)≠0The question on this page is asking for specific formula, which comes from Case 4, so I will only solve Case 4 here.Dividing numerator and denominator in (1) by cos(A)cos(B) we get:tan(A−B)=sin(A)/cos(B)−cos(A)/sin(B)/;2cosA)cos(B)/cos(A)cos(B)+sin(A)sin(B)/cos(A)cos(B)
cos(A)cos(B)tan(A−B)=sin(A)cos(B)/cos(A)cos(B)−cos(A)sin(B)cos/(A)cos(B)cos/(A)cos(B)co(A)cos(B)+sin(A)sin(B)/cos(A)cos(B)
cos(A)cos(B)tan(A−B)=sin(A)/cos(A)−sin(B)cos(B)1+sin(A)cos(A)sin(B)cos(B)
cos(A)−sin(B)cos(B)1+sin(A)cos(A)sin(B)cos(B) tan(A−B)=tan(A)−tan(B)/1+tan(A)tan(B)
Step-by-step explanation:
Hope it is helpful....
