MCQ
If $A + B + C = {180^o},$ then the value of $(\cot B + \cot C)$ $(\cot C + \cot A)\,\,(\cot A + \cot B)$ will be
- A$\sec A\,\sec B\,\sec C$
- ✓${\rm{cosec}}\,A\,{\rm{cosec}}\,B\,{\rm{cosec}}\,C$
- C$\tan A\,\tan B\,\tan C$
- D$1$
$ = \frac{{\sin (B + C)}}{{\sin B\,\sin C}}$
$ = \frac{{\sin ({{180}^o} - A)}}{{\sin B\,\sin C}}$
$ = \frac{{\sin A}}{{\sin B\sin C}}$
Similarly, $\cot C + \cot A = \frac{{\sin B}}{{\sin C\sin A}}$
and $\cot A + \cot B = \frac{{\sin C}}{{\sin A\sin B}}$
Therefore, $(\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B)$
$ = \frac{{\sin A}}{{\sin B\sin C}}.\frac{{\sin B}}{{\sin C\sin A}}.\frac{{\sin C}}{{\sin A\sin B}}$
$ = \cos {\rm{ec}}A\cos {\rm{ec}}B\cos {\rm{ec}}C.$
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$2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1, x \in[0, \pi]$
and $S$ is the sum of all these solutions, then the ordered pair $(\mathrm{n}, \mathrm{S})$ is :