How do you prove 4 (cos^4a + sin^4a) / (3+cos4a) = (cos^4a-sin^4a) sec2a?

Step-by-step explanation:

Step 1: (sin²A+cos²A)²

Step 2: (a+b)² = a²+b²+2ab

Step 3: (sin²A+cos²A)²=sin⁴+cos⁴+2sin²Acos²A

make sure to use: sin²a+cos²a=1

Step 4: (1)²-2sin²Acos²A = sin⁴A+cos⁴A

Step 5: 1-2sin²Acos²A = sin⁴A+cos⁴A

answerim :

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step-by-step explanation:

im not sure but my dad knows

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)