Can you solve this problem (SinA CosB + CosA SinB) / (CosA CosB) = TanA TanB using trigonometric identities?

3 ⤊

2007-03-07 01:10:39 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

Well, remember that SinQ / CosQ = TanQ. (You're welcome.)

Cancel like tems, and your expression reduces to TanA + TanB.

2007-03-07 01:15:32 · answer #1 · answered by etopro 2 · 1⤊ 0⤋

left hand side : (sinA cosB/CosA CosB)+(sinB cosA/CosA CosB)

i.e : sinA /CosA + sinB /CosB = R.H.S

2007-03-07 09:16:18 · answer #2 · answered by tins 2 · 1⤊ 1⤋

sinx/cosx=tanx

(sinAcosB + cosAsinB) / (cosAcosB) = tanA + tanB
(sinAcosB)/(cosAcosB) + (cosAsinB)/(cosAcosB) = tanA +tanB
tanA + tanB = tanA + tanB

2007-03-07 09:22:19 · answer #3 · answered by SensitiveMe 2 · 1⤊ 0⤋

Finally you arrive to
cot(A)+cot(B)=1
where cot(x) = 1/tan(x)

2007-03-07 09:31:12 · answer #4 · answered by 11:11 3 · 0⤊ 1⤋