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)

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