Question

Prove that sinA + sinB / sin A - sin B = tan(A+B/2). cot (A-B/2).
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Answer 1

Answer:

First we have cos(A+B)=cos(A)cos(B)−sin(A)sin(B) . Let’s divide by cos(A) since it is clearly not zero, and see what we get!

cos(A+B)/cos(A)=cos(A)cos(B)/cos(A)−sin(A)sin(B)/cos(A). Since tan(A)=sin(A)/cos(A)and according to the trigonometric identity, we also get cos2(B)+sin2(B)=1 which means that cos(B)=1/2–√ (plus or minus).

Thus we get cos(A+B)/cos(A)=cos(B)−tan(A)sin(B)=1/2–√−1×1/2–√=0, under the assumption that the cos(B)is positive. Since cos(A) is limited, cos(A+B) has to be zero, cool!

What if it is negative? Then we have cos(A+B)/cos(A)=−2/2–√=−2–√. A little bit harder. Now we can see that tan(A)=sin(A)/sin(A)=1 which means that sin(A)=cos(A). But we also have the trigonometric identity, hence this gives us cos(A)=1/2–√ (again plus or minus). But this would imply that cos(A+B) is larger than one (or smaller than -1), which is impossible!!! We are done!

Answer: cos(A+B)=0