A) Dependent on B
B) Dependent on A and B
C) Dependent on A
D) Independent of A and B
Correct Answer: C
Solution :
\[{{\cos }^{2}}(A-B)+{{\cos }^{2}}B-2\,\cos \,(A-B)\,\cos A\,\,\cos B\] \[={{\cos }^{2}}(A-B)+{{\cos }^{2}}B\]\[-\cos \,(A-B)\,\left\{ \cos (A-B)+\cos (A+B) \right\}\] \[={{\cos }^{2}}B-\cos \,(A-B)\,\,\cos \,\,(A+B)\] \[={{\cos }^{2}}B-({{\cos }^{2}}A-{{\sin }^{2}}B)=1-{{\cos }^{2}}A\] Hence it depends on A. Trick: Put two different values of A. Let \[A={{90}^{o}},\] then the value of expression will be \[{{\sin }^{2}}B+{{\cos }^{2}}B=1\] Now put \[A={{0}^{o}}\], then the value of expression will be \[{{\cos }^{2}}B+{{\cos }^{2}}B-2\,\,{{\cos }^{2}}B=0\] It means that the expression has different values for different A i.e. it depends on A. Now similarly for \[B={{90}^{o}},\] the value of expression will be \[{{\sin }^{2}}A+0-0\]\[={{\sin }^{2}}A\] and at \[B\,\,={{0}^{o}}\] the value of expression will be \[{{\cos }^{2}}A+1-2{{\cos }^{2}}A={{\sin }^{2}}A\]. Hence, the expression has the same value for different values of B, so it does not depend on B.
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