Question

Question 1 :
sin 8 A +sin 2 A/
cos 8 A +cos 2 A
....​

Answer 1

$\qquad\quad\boxed{\bf{\mid{\overline{\underline{\purple{\bigstar\: Correct~Question:-}}}}}\mid}$⠀

Prove that sin^8 A -cos^8 A =(sin^2- cos^2 A)(1-2 sin^2 A cos^2 A

$\qquad\quad\underline{\bf{\mid{\overline{\underline{\purple{\bigstar\: Answer:-}}}}}\mid}$⠀

starting with LHS :

=> sin⁸A - cos⁸A

we know that :-

sin⁸A = (sin⁴A)² 

cos⁸A = (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)²

cos⁴A = (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 - 2sin²A cos²A ]

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

Now ,

sin⁴A - cos⁴A 

this can be written in the form of the identity a² - b² = (a+b) (a -b)

sin⁴A  = (sin²A)²

cos⁴A = (cos²A)²

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

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

we know that ,

sin²A + cos²A = 1              [ by identity ]

hence,

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

=> ( 1 - 2sin²A cos²A ) (sin²A - cos²A)

=>RHS

Answer 2

Correct question➡️

⠀

Prove that sin^8 A -cos^8 A =(sin^2- cos^2 A)(1-2 sin^2 A cos^2 A

Answer➡️

⠀

starting with LHS :

=> sin⁸A - cos⁸A

we know that :-

sin⁸A = (sin⁴A)²

cos⁸A = (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)²

cos⁴A = (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 - 2sin²A cos²A ]

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

Now ,

sin⁴A - cos⁴A

this can be written in the form of the identity a² - b² = (a+b) (a -b)

sin⁴A = (sin²A)²

cos⁴A = (cos²A)²

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

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

we know that ,

sin²A + cos²A = 1 [ by identity ]

hence,

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

=> ( 1 - 2sin²A cos²A ) (sin²A - cos²A)

=>RHS

$\huge \pink {Hope\:it\:helps}$