If `(tan 3 A)/(tan A)=k,` prove that `(sin 3A)/(sin A)=(2 k)/(k-1)` and hence that deduce that either `k >3` or `k < 1/3.`

If (tan 3A)/(tan A)=k , show that ( sin 3A)/(sin A)= (2k)/(k-1) and hence or otherwise prove that either k gt 3 or k lt 1/3 .

If (tan 3A)/(tan A)=k , show that ( sin 3A)/(sin A)= (2k)/(k-1) and hence or otherwise prove that either k gt 3 or k lt 1/3 .

If (tan 3A)/(tan A)=k , show that ( sin 3A)/(sin A)= (2k)/(k-1) and hence or otherwise prove that either k gt 3 or k lt 1/3 .

If (tan3A)/(tan A)=k, then (sin3A)/(sin A)=

"If" (tan 3 A)/(tan A) = k ", show that" ( sin 3 A)/(sin A) = (2k)/(k-1), also show that the value of k cannot lie between 1/3 "and" 3

If (tan3A)/(tanA)=k, then (sin3A)/(sinA)=

If tan3A=k tanA , Then S.T (sin3A)/(sinA)=(2K)/(K-1)

If tan 3A /tan A =k , show that sin 3A/sin A = 2k/k-1 an and k cannot Second Method lie between 1/3 and 3.