sinAcosB + cosAsinB tanA + tanB
------------------------------ = ------------------
cosAcosB - sinAsinB 1-tanAtanB
2007-12-05 02:00:09 · 3 answers · asked by Singh J 1 in Science & Mathematics ➔ Mathematics
The RHS should alert you that you want to get the LHS in the form tan(A+B) = sin(A+B) / cos(A+B)
2007-12-05 02:07:59 · answer #1 · answered by Dr D 7 · 0⤊ 0⤋
(sin A cos B + cos A sin B)/(cos A cos B - sin A sin B)
Divide both numerator and denominator by cos A cos B:
(sin A/cos A+ sin B/cos B)/(1 - sin A sin B/(cos A cos B))
Simplify:
(tan A + tan B)/(1 - tan A tan B)
And we are done.
2007-12-05 10:08:56 · answer #2 · answered by Pascal 7 · 0⤊ 0⤋
you know we did this sum today in the class but i just can't remember it ......................................................... pls frwrd othr question
2007-12-05 10:10:13 · answer #3 · answered by Lara G 2 · 0⤊ 0⤋