Evaluate!

Let $a=\cos \theta$ , $b=\sin \theta$ , $c=\cos \beta$ and $d=\sin \beta$ .

Substituting in the third equation, we obtain:

$\cos \theta \cos \beta+\sin \theta \sin \beta=0$

$\cos(\theta-\beta)=0$

Finally, let $x=ab+cd$ , so:

$x=\cos \theta \sin \theta+\cos \beta \sin \beta$

$x=\dfrac{1}{2}\left(\sin 2\theta+\sin 2\beta\right)$

Use the sum to product formulas:

$x=\sin(\theta+\beta)\cos(\theta-\beta)$

$x=\sin(\theta+\beta)(0)$

$x=\boxed{0}$

$ac + bd = 0$

$ac=-bd$

Squaring both sides

$a^2 c^2 = b^2 d^2$ -equation p)

Since $b^2 = 1 - a^2, d^2 = 1 - c^2$

So putting in equation p)

$a^2 c^2 = (1-a^2)(1-c^2)$

$a^2 c^2 = 1 + a^2 c^2 -a^2 - c^2$

$a^2 + c^2 = 1$

But $a^2 + b^2 =1$

Hence $b^2 = c^2$ Equation q)

Also $c^2 + d^2 =1$

Hence $a^2 = d^2$ Equation r)

We need to find (ab+cd)

so squaring

$(ab+cd)^2=a^2 b^2 + c^2 d^2 + 2abcd$

$=a^2 c^2 + c^2 a^2 + 2abcd$ Using q) and r)

$=2a^2 c^2 + 2abcd$

$=2ac(ac+bd)$

$=2ac*0$

$=0$

Since $(ab+cd)^2 = 0$

Hence ab+cd=0.

We know (a+b)^2 = a^2 + b^2 + 2ab and (a-b)^2= a^2 + b^2 - 2ab

Using the information given in the problem we deduce

(a+b)^2 = 1+ 2ab Similarly (c+d)^2 = 1+ 2cd

Multiplying the above two eqns we have

(a+b)^2(c+d)^2 = (1+2ab)(1+2cd)

[(a+b)(c+d)]^2 = 1+ 4abcd + 2(ab+cd)

(ad+bc)^2 = 1 + 4abcd + 2(ab+ cd) …… since given that ac+bd=0

(ad+bc)^2 = 1 + 4abcd + 2(ab+ cd) we will name this eqn 1

Again using the information given in the problem we have

(a-b)^2 = 1-2ab and (c-d)^2 = 1- 2cd

Multiplying the above two eqns we get

(a-b)^2(c-d)^2 = (1-2ab)(1-2cd)

[(a-b)(c-d)]^2 = 1+ 4abcd - 2(ab+cd)

[- (ad+bc)]^2 = 1+ 4abcd - 2(ab+cd) …… since given that ac+bd=0

Or (ad+bc)^2 = 1+ 4abcd - 2(ab+cd) we will name this eqn 2

Eqn 1 and eqn 2 represent the same value.

Hence equating them , we get

1 + 4abcd + 2(ab+ cd) = 1+ 4abcd - 2(ab+cd)

Or 4(ab+cd) = 0

And so ab+cd= 0

a,b,c,d can be seen as coordinates of two points A(a.b) and C(c,d) on 2D Cartesian plane. Both points are on the unit circle since both sums of square are equal to 1.
ac+bd=0 indicates segment OA and OC are perpendicular where O is the origin of the plane. For a shortcut, take A as (1,0) and C as (0,1) , ab + cd = 0.

The easiest of the ways is substitution. Let us assume a=c=1 and b=d=0 . Since this assumption satisfies all the conditions mentioned in the question ,it is correct. Therefore, ab+cd = (1)(0) + (1)(0) = 0

$x=ab+cd$

$x^2=(a^2-d^2)(b^2-c^2)$

$We have a^2+b^2-c^2-d^2=0 => a^2-d^2=-(b^2-c^2)$

$Thus x^2=-(a^2-d^2)^2 => x=0$

ac+bd=0.

ac= -bd

a/b = -d/c sqare both sides

a^2/b^2 =d^2/c^2 apply componendo

property of proportion

a^2+b^2/b^2= d^2+c^2/c^2

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

1/b^2= 1/c^2

b^2= c^2 implies b= c

So ab+cd =ac+-bd=0

Awswering Ashley Shamidha. When you see the sum of two real quadratic numbers, stay smart to utilize this technique.

Notice that if $a=1$ , $b=0$ , $c=0$ , and $d=1$ , then all of the equations will hold true. Because there is only one numerical answer to the problem, the answer must be $1\cdot0+0\cdot1=\boxed{0}$ .