I+sinx.siny.sin A cos A (sin2 A- cos2 A) (1- 2 sin2 A. coscot4 A + cot

1.	I+sinx.siny.sin A cos A (sin2 A- cos2 A) (1- 2 sin2 A. coscot4 A + cot2 A = cosec 4A-coset,2A2 sec 2 A-sec 4 A-2 cosec2A + cosec4A_ cot4 A-ta(sin A + cosec A)2 + (cos A+ sec A)2= tan2 A+ cota(1 + cot A-cosec A) (1 + tan A+ sec A) = 2)
Answer» starting with LHS : =>sin⁸A - cos⁸A we know that :-sin⁸A = (sin⁴A)²cos⁸A = (cos⁴A)² This can be written as :- =>(sin⁴A)² - (cos⁴A)² Now this is in the form of an identity : a² - b² = (a+b) ( a - b) =>(sin⁴A + cos⁴A) ( sin⁴A - cos⁴A) sin⁴A = (sin²A)²cos⁴A = (cos²A)² =>(sin²A)² +(cos²A)² (( sin⁴A - cos⁴A)) (sin²A)² +(cos²A)² can be in the identity : a² + b² = (a+b)² - 2ab [ (sin²A)² +(cos²A)² = (sin²A + cos²A)² - 2sin²Acos²A ] => [(sin²A + cos²A)² - 2sin²Acos²A ] (( sin⁴A - cos⁴A)) Now ,sin⁴A - cos⁴Athis can be written in the form of the identity a² - b² = (a+b) (a -b) sin⁴A = (sin²A)²cos⁴A = (cos²A)² sin⁴A - cos⁴A = (sin²A + cos²A) (sin²A - cos²A) =>=> [(sin²A + cos²A)² - 2sin²Acos²A ](sin²A + cos²A) (sin²A - cos²A) we know that ,sin²A + cos²A = 1 [ by identity ] hence, => [(1)²- 2sin²Acos²A ](1)×(sin²A - cos²A) => ( 1 -2sin²Acos²A ) (sin²A - cos²A) =>RHS Thank you Question thik se post karo

I+sinx.siny.sin A cos A (sin2 A- cos2 A) (1- 2 sin2 A. coscot4 A + cot2 A = cosec 4A-coset,2A2 sec 2 A-sec 4 A-2 cosec2A + cosec4A_ cot4 A-ta(sin A + cosec A)2 + (cos A+ sec A)2= tan2 A+ cota(1 + cot A-cosec A) (1 + tan A+ sec A) = 2)

Answer»

starting with LHS : =>sin⁸A - cos⁸A

we know that :-sin⁸A = (sin⁴A)²cos⁸A = (cos⁴A)²

This can be written as :-

=>(sin⁴A)² - (cos⁴A)²

Now this is in the form of an identity : a² - b² = (a+b) ( a - b)

=>(sin⁴A + cos⁴A) ( sin⁴A - cos⁴A)

sin⁴A = (sin²A)²cos⁴A = (cos²A)²

=>(sin²A)² +(cos²A)² (( sin⁴A - cos⁴A))

(sin²A)² +(cos²A)² can be in the identity : a² + b² = (a+b)² - 2ab

[ (sin²A)² +(cos²A)² = (sin²A + cos²A)² - 2sin²Acos²A ]

=> [(sin²A + cos²A)² - 2sin²Acos²A ] (( sin⁴A - cos⁴A))

Now ,sin⁴A - cos⁴Athis can be written in the form of the identity a² - b² = (a+b) (a -b)

sin⁴A = (sin²A)²cos⁴A = (cos²A)²

sin⁴A - cos⁴A = (sin²A + cos²A) (sin²A - cos²A)

=>=> [(sin²A + cos²A)² - 2sin²Acos²A ](sin²A + cos²A) (sin²A - cos²A)

we know that ,sin²A + cos²A = 1 [ by identity ]

hence,

=> [(1)²- 2sin²Acos²A ](1)×(sin²A - cos²A)

=> ( 1 -2sin²Acos²A ) (sin²A - cos²A)

=>RHS

Thank you

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