If CosA = mCosB, then prove that :-
No rush for points, please answer with good explanations.
Answers
Answered by
38
using coponendo dividendo ie if a/b = c/d then (a+b)/(a-b)= (c+d)/(c-d)
so (CosA + CosB)/(CosA - CosB)=(m +1)/(m -1)
using formulas
[2Cos {(A + B)/2}Cos{(B - A)/2}]/[2Sin{(B- A)/2}Sin{(A+ B )/2}]=(m +1)/(m -1)
do cross multiplication
2 will be canceled
Cos{(A + B)/2}/Sin{(A + B)/2}={(m +1)/(m -1)} × Sin{(B - A)/2}/Cos {(B - A)/2}
we know that cosθ/sinθ = cotθ and sinθ/cosθ= tanθ
so we get
Cot{(A + B)/2}={(m +1)/(m -1)} Tan{(B - A)/2}
PROVED
Anonymous:
do you got it ok then so you understood the last step of cross muliplication
ok plz mark as best and say thank you
means tick the thank you tab
Wait bro, I will work on your solution
got to go, thanks, and be sure, I will mark your answer as best, just if you can further the steps in your answer, it will be easy for me to understand
Oh nice, I didn't know there were such formulas too ! Thanks !
http://www.maths.manchester.ac.uk/~cds/internal/tables/trig.pdf
Look here, the formula is different. It shows A-B in the RHS, but in your answer, it is B-A, can you tell why ?
Look here, the formula is different. It shows A-B in the RHS, but in your answer, it is B-A, can you tell why ?
and after expanding CosA-CosB, there should be a negative sign on one of the terms of RHS, where is that in your answer ?
thankx
Answered by
0
Step-by-step explanation:
Hope it helps!!!
best of luck
Attachments:
Similar questions