prove that sin4a/cos2a*1-cos2a/1-cos4a=tana

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prove that sin4a/cos2a*1-cos2a/1-cos4a=tana

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Mackenzie 1 month 2021-08-17T18:10:33+00:00 1 Answer 0 views 0

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    2021-08-17T18:12:03+00:00

    Answer:

    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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