A) A
B) B
C) both A and B
D) neither A nor B
Correct Answer: B
Solution :
\[{{\cos }^{2}}(A-B)+oc{{s}^{2}}B-2\cos (A-B)cos\,A\,cos\,B\] \[={{\cos }^{2}}(A-B)+co{{s}^{2}}B-\cos (A-B)\] \[[cos(A+B)+cos(A-B)]\] \[={{\cos }^{2}}B-\cos (A-B)cos(A+B)\] \[={{\cos }^{2}}B-\frac{1}{2}[cos2A+cos2B]\] \[={{\cos }^{2}}B-\frac{1}{2}[2co{{s}^{2}}B-1+cos2A]\] \[=\frac{1}{2}-\frac{1}{2}\cos 2A\] \[\frac{1}{2}-\frac{1}{2}(2co{{s}^{2}}A-1)\] \[=1-{{\cos }^{2}}A={{\sin }^{2}}A\] Hence, it is independent of B.
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