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sin4x+cos4x=1-2sin2xcos2x

LHS- sin4x + cos4x

= (sin2x)2 + (cos2x)2

= (sin2x + cos2x)2 - 2sin2xcos2x  {since a2 +b2 = (a+b)2 - 2ab}

= 12 - 2sin2xcos2x {since sin2x +cos2x = 1}

= 1 - 2sin2xcos2x

= RHS

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             LHS =sin4 X +cos4X 

                     =(sin2X + cos2X)2 - 2 sin2X cos2X          [ a2 + b2 = (a + b )2 - 2ab ]

                     =1 - 2 sin2X cos2X

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Here you go

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sin²A. tan²A+cos²A.cot²A=?
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i don't understand
 
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Cos²A. Tan²A + cos²A.cot²A

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ohh thanku for all for these question i m completed my homework ​
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Please find your solution below

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Please find your solution below

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hope both the answers will help you
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we know that,
    sin2x+cos2x=1
or (sin2x+cos2x)2=1
or sin4x+cos4x+ 2.sin2x.cos2x=1
or sin4x+cos4x=1-2.sin2x.cos2x
                                            Proved
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Sorry for the quality of paper


Hope you benefitted
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H.P
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Sorry, another answer was posted by mistake
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