If sinA=ksinB, Prove that tan(1/2)(A-B)=(k-1)/(k+1)tan(1/2)(A+B)

Steps to prove if sinA=ksinB, Prove that tan(1/2)(A-B)=(k-1)/(k+1)tan(1/2)(A+B)

From Question given that

⇒sinA=ksinB

⇒sinA/sinB=k/1

⇒(sinA+sinB)/(sinA-sinB)=(K+1)/(k-1)

⇒{2.sin(A+B)/2.cos(A-B)/2}/{2.cos(A+B)/2.cos(A-B)/2}=(K+1)/(k-1)

⇒tan(A+B)/2.cot(A-B)/2=(K+1)/(k-1)

⇒(K-1)/(k+1)tan(A+B)/2=1/cot(A-B)/2

⇒tan(1/2)(A-B)=(k-1)/(k+1)tan(1/2)(A+B)

Hence tan(1/2)(A-B)=(k-1)/(k+1)tan(1/2)(A+B) is proved.

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