Question

please answer soon iam I have exams​

Answer 1

Step-by-step explanation:

$\frac{SinA-2Sin^{3}A }{2Cos^{3}A-CosA}\\ =\frac{SinA(1-2Sin^{2}A )}{CosA(2Cos^{2}A-1)}\\ =\frac{SinA(Sin^{2}A+Cos^{2}A-2Sin^{2}A)}{CosA(2Cos^{2}A-Sin^{2}A-Cos^{2}A)}\\ =\frac{SinA(Cos^{2}A-Sin^{2}A)}{CosA(Cos^{2}A-Sin^{2}A)}\\ =\frac{SinA}{CosA}\\ =TanA$

(Hence Proved)