Question

The expression ${\cos }^2(A-B)+{\cos }^{2}B-2\cos (A-B)\cos A\cos B$ is independent of

• A. A
• B. B
• C. both A and B
• D. neither A nor B

Answer 1

Answer: (B)

${\cos }^2(A-B)+{\cos }^{2}B-2\cos (A-B)\cos A\cos B$
$={\cos }^2(A-B)+{\cos }^{2}B-\cos (A-B)$
$[\cos (A+B)+\cos (A-B)]$
$={\cos }^{2}B-\cos (A-B)\cos (A+B)$
$={\cos }^{2}B-\frac{1}{2}[\cos 2A+\cos 2B]$
$={\cos }^{2}B-\frac{1}{2}\left[2{\cos }^{2}B-1+\cos 2A\right]$
$=\frac{1}{2}-\frac{1}{2}\cos 2A$
$=\frac{1}{2}-\frac{1}{2}\left(2{\cos }^{2}A-1\right)$
$=1-{\cos }^{2}A={\sin }^{2}A$
Hence, it is independent of $B$.