solve it again please
Answers
After doing the question, I suggest there might be a misprint in to prove. I think it should be β instead of α. Pls check, as u can see what I've tried to do in the Q. Inform me after checking.
Let,
α=a, β=b to avoid confusions
tanA = Q sinB/(P + Q cosB)
TP- tan(B-A) = P sinB/ (Q + P cosB)
LHS: tan(B-A) = (tanB - tanA) / 1 + tanA tanB
I will calculate them separately too avoid confusions
- tanB - tanA = sinB/cosB - Q sinB/(P + Q cosB)
= (P sinB + Q sinB cosB - Q sinB cosB) / cosB (P + Q cosB)
= P sinB / cosB (P + Q cosB)
- 1 + tanA tanB = 1 + sinB/cosB * Q sinB/(P + Q cosB)
= P cosB + Q cos²B + Q sin²b/ cosB (P + Q cosB)
= P cosB + Q (sin²b + cos²B)/ cosB (P + Q cosB)
= Q + P cosB/ cosB (P + Q cosB)
Returning to the original equation
= {P sinB / cosB (P + Q cosB)} / {Q + P cosB/ cosB (P + Q cosB)}
= P sinB / (Q + P cosB) = RHS
Hope I helped
Extra:
If we try to take the value of SinB from given
SinB = (Q + P cosB) sinA / Q cosA
After substituting this in what we got after simplifying tan(B-A)
= P (Q + P cosB) sinA / Q cosA (Q + P cosB)
= P sinA/ Q cosA
I don't think I would be able to change this into the term asked by the original Q, so I suggested the correction.