Prove that sin3a-cos3a/sina+cosa=2sin2a-1

To prove : sin(3a)-cos(3a)/sin(a)+cos(a) = 2sin(2a)-1                                      

Step-by-step explanation:

\frac{\sin3a-\cos3a}{\sin a+\cos a}=2\sin 2a-1\\\cos3a=4\cos^{3}a-3\cos a\\\sin 3a=3\sin a-4\sin ^{3}a\\substituting the value, we get\\L.H.S= \frac{3\sin a-4\sin ^{3}a-[4\cos ^{3}a-3\cos a]}{\sin a+\cos a}\\\frac{3(\sin a+\cos a)-4(\sin ^{3}a+4\cos ^{3}a)}{\sin a+\cos a}\\\frac{3(\sin a+\cos a)-4[(\cos a+\sin a)(\cos ^{2}a+\sin ^{2}a-\sin a\cos a)]}{\sin a+\cos a}\\\frac{3(\sin a+\cos a)-4[(\cos a+\sin a)(1-\sin a\cos a)]}{\sin a+\cos a}\\\frac{(\sin a+\cos a)[3-4(1-\sin a\cos a)]}{\sin a+\cos a}\\\frac{(\sin a+\cos a)[3-4+4\sin a\cos a]}{\sin a+\cos a}\\\frac{(\sin a+\cos a)(-1+4\sin a\cos a)}{\sin a+\cos a}\\-1+4\sin a\cos a\\-1+2\sin 2a\\2\sin 2a-1=R.H.S

Since, sin(2a)= 2sin(a)cos(a)

2sin(2a)= 4sin(a)cos(a)

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step-by-step explanation:

1+2+3+4+5+6+7+8+9+10 = 55

2*10 = 20

55-20 = 35