prove that sin4a/cos2a*1-cos2a/1-cos4a=tana
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Step-by-step explanation:LHS = cos^6 A + sin^6 A
= (cos²A)^3 + (sin²A)^3
Using identity (a + b)^3 = a^3 +b^3 + 3ab ( a+b)
=> a^3 +b^3 = ( a+ b)^3 -3ab ( a+ b)
So, (cos²A)^3 + ( sin² A)^3
=(cos²A+ sin²A)^3 - 3 cos²A sin²A(cos²A+sin² A)
= 1- 3cos² A sin² A ( since sin² A + cos² A = 1 ) ……………….. LHS
Now, RHS = { 1 + 3cos² ( 2A)} / 4
= { 1 + 3 ( cos 2A )² } /4
= {1+ 3 ( cos² A - sin² A )² } /4 { since cos 2A = cos² A - sin² A)
= { 1+ 3( cosA + sinA)² ( cosA - sinA)² }/4
= {1+ 3( 1 + 2sinA cosA) ( 1 - 2sinA cosA)} /4
= { 1 + 3 ( 1 - 4 sin² A cos² A) } / 4
= { 1 + 3 - 12 sin² A cos² A} /4
= (4 - 12 sin² A cos² A) /4
=> 1 - 3 sin²A cos ² A ………….. RHS
Hence proved that LHS = RHS
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