find the general solution of sin2AsecA + 31/2tanA =0 Share with your friends Share 0 Varun.Rawat answered this The given equation is, sin2A . sec A + 3 tan A = 0⇒sin Acos A . sin A + 3 tan A = 0⇒tan A . sin A + 3 tan A = 0⇒tan Asin A + 3 = 0⇒tan A = 0 or sin A + 3 = 0⇒tan A = 0 or sin A = -3 rejected as -1≤sin θ≤1⇒tan A = 0⇒A = nπ , where n ∈Z -1 View Full Answer Raghav answered this HOPE IT HELPS!!!! Vote it helpful. 2 Matin Saiyed answered this thanks 0
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