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.

"> 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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If \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\]are in A. P. then which of the following are also in A.P. [ISM Dhandbad 1989]

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 10 months ago

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\[{{\SIN }^{2}}B-{{\sin }^{2}}A={{\sin }^{2}}C-{{\sin }^{2}}B\] \ \[\sin (B+A)\sin (B-A)=\sin (C+B)\sin (C-B)\] 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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