ABCD's are driving you crazy

$\Rightarrow(ad-bc)^2+(ac+bd)^2$ $\Rightarrow a^2d^2-2abcd+b^2c^2+a^2c^2+2abcd+b^2d^2$ - Cancel the 2abcd. $\Rightarrow$ $a^2d^2+b^2d^2+a^2c^2+b^2c^2$ $\Rightarrow d^2(a^2+b^2)+c^2(a^2+b^2)$ $\Rightarrow (a^2+b^2)(c^2+d^2)$ - Since $a^2+b^2=2$ and $c^2+d^2=1$ , we have: $2\cdot 1=2$

In order for $a^2 + b^2$ to be 2 , a and b would most likely have to be 1 .

$1^2 + 1^2 = 1$

The second one can be true if c is 1 and d is 0 (or vice versa)

$1^2 + 0^2 = 1$

So far:

a = 1

b = 1

c = 1

d = 0

Plug in the values into the last equation and you get:

$(1 * 0 - 1 * 1)^2 + ( 1 * 1 + 1 * 0 )^2 = 2$

It is well-known that $(a^2 + b^2) ( c^2 + d^2) = (ad - bc)^2 + (ac+bd)^2 = \boxed{2}$

This is Lagrange's Identity.

Given a^2 + b^2 = 2 c^2+d^2 = 1

(ad-bc)^2+(ac+bd)^2 = a^2d^2+b^2c^2+a^2c^2+b^2d^2 =(c^2+d^2)(a^2+b^2) =2

Answer. .. it's. simple. algebra.

CHALLENGE: Use trigonometry in this question.

Its just simple lagrange's identity

Take a=√2 sinx, b=√2cosx, c=sinx, d=cosx. You will get your answer...

Let $x=a-bi$ , $y=c+di$ .
$|x|^2=2$ , $|y|^2=1$
$(ad-bc)^2+(ac+bd)^2=|xy|^2=|x|^2|y|^2=2\times 1=2$

Notice that (a^2+b^2)(c^2+d^2)=(ad-bc)^2+(ac+bd)^2. This can be formulated just by multiplying the terms.

Hence the solution is (a^2+b^2)(c^2+d^2)=2 [provided]

Same as Mohammad Saleem !