Trig prob

\frac{tan 3A}{tan A}=\frac{3 tanA - tan^{3}A}{1- 3 tan^{2}A}.\frac{1}{tan A} = k

\left(1- 3 tan^{2}A \right)(k) tan A = 3 tan A - tan ^{3}A

(k) tan A - (3k) tan^{3}A = 3 tanA - tan^{3}A

k \tan A - 3 \tan A = 3k \tan^{3}A - tan^{3}A

(k-1) = \tan ^{2}A (3k-1)

Therefore, we get
tan^{2}A = \frac{k-1}{3k-1}

Now since, \frac{Sin\3A}{Sin A} = \frac{3 Sin\ A - 4 Sin^{3}A }{Sin\ A}

= 3 - 4\ Sin^{2}\ A

= 3 - \frac{4}{1 + \frac{1}{tan ^{2}\ A}}

= 2kk-1

Therefore, (a) option

wonderful!
amazing!

Thank you.

well i think i have a better solution maybe...

tan3AtanA = k

sin3A.cosAsinA.cos3A = 2sin3A.cosA2sinA.cos3A = sin4A + sin2Asin4A - sin2A = k

=> sin4Asin2A = (k + 1)(k - 1)

=> 2cos2A = (k + 1)(k - 1)

Now, sin3AsinA = 3 - 4sin²A

= 3 - 2(1 - cos2A) = 3 - 2[1 - (k + 1)2(k - 1)] where [] is not G.I.F ... :P lol

= 3 - 2{(k - 3)2(k - 1)}

= 3(k - 1) - (k - 3)(k - 1)

= 2k(k - 1) done..!!