In a quadrilateral ABCD, cosAcosB + sinCsinD =
$

  (a){\text{ cosCcosD + sinAsinB}} \\

  (b){\text{ cosCcosD - sinAsinB}} \\

  (c){\text{ sinCsinD - cosAcosB}} \\

  (d){\text{ sinA + sinB + sinC + sinD}} \\

 $

Hint: In this question use the concept that the sum of opposite angles of a quadrilateral is ${180^0}$. Thus A+C=180 and B+D=180. Thus use these two equations and apply cosine to get the answer.

Complete step-by-step answer:


As we know that in a cyclic quadrilateral the sum of opposite angles is 180 degrees.
So let us assume a quadrilateral ABCD as shown in the figure whose vertices are taken in order.

Therefore the sum of angles A and C and B and D should be 180 degrees.

$ \Rightarrow A + C = {180^0}$....................... (1)

$ \Rightarrow B + D = {180^0}$ .................. (2)

Now from equation (1) and (2) we can say that

$ \Rightarrow A + C = B + D$

$ \Rightarrow A - B = D - C$

Now apply cosine on both sides we have,

$ \Rightarrow \cos \left( {A - B} \right) = \cos \left( {D - C} \right)$

Now expand the above equation according to cosine property which is $\left( {\cos \left( {A -
B} \right) = \cos A\cos B + \sin A\sin B} \right)$ we have,

$ \Rightarrow \cos A\cos B + \sin A\sin B = \cos D\cos C + \sin D\sin C$

$ \Rightarrow \cos A\cos B - \sin C\sin D = \cos C\cos D - \sin A\sin B$

So this is the required answer.

Hence option (B) is correct.

Note: A quadrilateral is a 4 sided plane figure having 4 vertices and 4 angles. The total sum of its interior angles is ${360^0}$. The tricky part here was the applying of cosine, as the equations were yielding us sides and we wished to have sides within cosine that’s why we took it. It is advised to remember direct trigonometric identities like $\left( {\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B} \right)$.