Question

LEVEL - 11
In a quadrilateral ABCD, COSA.cosB + sin sinD =
a) cosC cosD + sinA sinB
b) cosC cosD - sinA sinB
c) sinC sinD - COSA cosB
d) sinA + sinB + sinC + sind​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{ABCD is a quadrilateral}$

$\underline{\textsf{To find:}}$

$\textsf{The value of}$

$\mathsf{cosA\;cosB+sinC\;sinD}$

$\underline{\textsf{Solution:}}$

$\textsf{Since ABCD is a quadrilatel, we have}$

$\mathsf{A+B+C+D=360^{\circ}}$

$\mathsf{Consider,}$

$\mathsf{cosA\;cosB+sinC\;sinD}$

$\textsf{Using the identity}$

$\boxed{\mathsf{cos(A+B)=cosA\,cosB-sinA\,sinB}}$

$\mathsf{=cos(A+B)+sinA\,sinB+cosC\,cosD-cos(C+D)}$

$\mathsf{=cos(A+B)-cos(C+D)+sinA\,sinB+cosC\,cosD}$

$\mathsf{=cos(360^{\circ}-(C+D))-cos(C+D)+sinA\,sinB+cosC\,cosD}$

$\mathsf{=cos(C+D)-cos(C+D)+sinA\,sinB+cosC\,cosD}$

$\mathsf{=sinA\,sinB+cosC\,cosD}$

$\implies\boxed{\mathsf{cosA\;cosB+sinC\;sinD=sinA\,sinB+cosC\,cosD}}$

$\underline{\textsf{Answer:}}$

$\textsf{Option (a) is correct}$