Grade 11Trigonometry
If cosecA + secA = cosecB+secB, prove that tanA tanB tan{(A+B)/2} = 1
If cosecA + secA = cosecB+secB, prove that tanA tanB tan{(A+B)/2} = 1
1 Answer
Arun
Dear student
cosecA+secA=cosecB+secB
i.e cosecA-cosecB=secB-secA
=1/sinA-1/sinB=1/cosB-1/cosA
=(sinB-sinA)/sinAsinB=(cosA-cosB)/cosAcosB
=(2sin((B-A)/2)cos((B+A)/2))/sinAsinB=(2sin((B-A)/2)sin((B+A)/2))/cosAcosB
NOTING DOWN THE CONDITION ONLY IF A IS NOT EQUAL TO B CANCEL sin(B-A)/2 from both sides and cross the products to obtain
cos((b+a)/2)/sin((a+b)/2)=sinAsinB/cosAcosB
i.e cot(A+B)/2=tanAtanB
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