Question

if A+B+C=π and sin(A+C/2)=nsinC/2 , then show that tanA/2 tanB/2=n-1/n+1​
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Answer 1

Answer:

sin(A+C/2)=sin{A+(π/2-A/2-B/2)

=sin(π/2+A/2-B/2)=cos(A/2-B/2)

sinC/2=sin(π/2-A/2-B/2)=cos(A/2+B/2)

Since, sin(A+C/2)=ksinC/2,

so cos(A/2-B/2)= kcos(A/2+B/2)

cosA/2cosB/2+sinA/2sinB/2=kcosA/2cosB/2-ksinA/2sinB/2

1+tanA/2tanB/2=k-ktanA/2tanB/2

tanA/2tanB/2=(k-1)/(k+1).

Step-by-step explanation:

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