If (3--tan^2A)/(1--3tan^2A) = k. then find cosecA(3sinA-- 4sin^3A) in terms of k Share with your friends Share 1 Manbar Singh answered this We have, 3 - tan2A1 - 3 tan2A = k⇒3 tan A - tan3Atan A1 - 3 tan2A = k⇒3 tan A - tan3A1 - 3 tan2Atan A = k⇒tan 3Atan A = k ....1Now, tan 3Atan A - 1 = k - 1⇒tan 3A - tan Atan A = k- 1⇒sin 3Acos 3A - sin Acos Asin Acos A = k - 1⇒sin 3A . cos A - cos 3A . sin Acos A . cos 3A × cos Asin A = k - 1⇒sin3A - Asin A . cos 3A = k - 1⇒sin 2Asin A . cos 3A = k - 1⇒2 sin A . cos Asin A . cos 3A = k - 1⇒2 cos Acos 3A = k - 1 .....2Dividing 2 by 1, we get 2 cos Acos 3A ×tan Atan 3A = k - 1k⇒2 cos Acos 3A × sin Acos A × cos 3Asin 3A = k - 1k⇒sin Asin 3A = k - 12k⇒sin 3Asin A = 2kk - 1⇒cosec A 3 sin A - 4 sin3A = 2kk - 1 -4 View Full Answer