A) \[\sin A,\sin B,\sin C\]
B) \[\tan A,\tan B,\tan C\]
C) \[\cot A,\cot B,\cot C\]
D) None of these
Correct Answer: C
Solution :
\[{{\sin }^{2}}B-{{\sin }^{2}}A={{\sin }^{2}}C-{{\sin }^{2}}B\] \ \[\sin (B+A)\sin (B-A)=\sin (C+B)\sin (C-B)\] or \[\sin C(\sin B\cos A-\cos B\sin A)\] \[=\sin A(\sin C\cos B-\cos C\sin B)\] Divide by \[\sin A\sin B\sin C\] \[\therefore \,\,\,\cot A-\cot B=\cot B-\cot C\]. Hence the result.
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