Question

In A ABC angleC = 90 degree TanA=1÷√3

find
i) SinA CosB
plus CosA SinB
ii) CosA Cos B
SinA SinB.
​

Answer 1

Answer:

in ∆ABC angle C=90° tanA=1/√3=tan30°

A =30° B=60° ,so A+B=90°

Now SinACosB+CosAsinB=Sin(A+B)=Sin90°=1.

CosACosB +SinASinB=Cos(A-B)=Cos(B-A)=Cos(60-30)=Cos30°=√3/2