Let m 1 , m 2 be the slopes of two adjacent sides of a square of side a such that a 2 + 11 a + 3 m 1 2 + m 2 2 = 220 . If one vertex of the square is 10 cos α - sin α , 10 sin α + cos α , where α ∈ 0 , π 2 and the equation of one diagonal is cosα - sinα x + sin α + cos α y = 10 , then 72 sin 4 α + cos 4 α + a 2 - 3 a + 13 is equal to

Step-by-step solution

Given m 1   &   m 2 are slopes of two adjacent sides of square then, m 1 m 2 = - 1 Also given a 2 + 11 a + 3 m 1 2 + m 2 2 = 220 ⇒ a 2 + 11 a + 3 m 1 2 + 1 m 1 2 = 220 Now on plotting the diagram we get, Now given equation of AC , cos α - sin α x + sin α + cos α y = 10 Now by property of square we know that diagonals are perpendicular to each other So equation of B D will be sin α + cos α x + sin α - cos α y + λ = 0 now is passes through D 10 cos &#94