MCQ
${d \over {dx}}\left( {{{{{\cot }^2}x - 1} \over {{{\cot }^2}x + 1}}} \right) = $
- A$ - \sin 2x$
- B$2\sin 2x$
- C$2\cos 2x$
- ✓$ - 2\sin 2x$
$ = \frac{d}{{dx}}[\cos 2x] = - 2\sin 2x$.
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$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$
and $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ then