Question

If cosA = ncosB and sinA = msinB then show that (m^2-n^2) sinB = 1-n^2
​

Answer 1

Answer:

Answer is give in attachment

Thanks

Answer 2

$\huge \tt { \underline{ \underline{ᗩnswer}}}$

• LHS=(m²-n²)Sin²A
• =[(Sin²Α/Sin²B)-(Cos²Α/Cos²Β)](Sin²B)
• = [(Sin²Α Cos²Β-Cos²Α Sin²Β)/Sin²Β Cos²Β] *Sin²B
• =( Sin²Α Cos²Β/Cos²B)-(Cos²Α Sin²B/Cos²Β)
• = Sin²Α- Cos²Α(1-Cos²Β)/Cos²Β
• = Sin²Α- (Cos²Α/Cos²Β)+ Cos²Α Cos²Β/Cos²B
• = Sin²Α+Cos²Α- (Cos²Α/Cos²B)
• = 1- n²