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Question Number 108950 by ZiYangLee last updated on 20/Aug/20

$$\mathrm{In}△\mathrm{ABC},$$$$\mathrm{prove}\mathrm{that}\frac{1+\mathrm{cosA}-\mathrm{cosB}+\mathrm{cosC}}{1+\mathrm{cosA}+\mathrm{cosB}-\mathrm{cosC}}=\tan \frac{\mathrm{B}}{2}\cot \frac{\mathrm{C}}{2}$$

Answered by mnjuly1970 last updated on 20/Aug/20

$$\frac{(1+cos\left(\mathrm{C}\right))+(cos\left(\mathrm{A}\right)-cos\left(\mathrm{B}\right))}{(1-cos\left(\mathrm{C}\right))+(cos\left(\mathrm{A}\right)+cos\left(\mathrm{B}\right))}$$$$=\frac{2{cos}^2\left(\frac{\mathrm{C}}{2}\right)-2sin\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)sin\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}{2{sin}^2\left(\frac{\mathrm{C}}{2}\right)+2cos\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)cos\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}$$$$=\frac{2cos\left(\frac{\mathrm{C}}{2}\right)[sin\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)-sin\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)]}{2sin\left(\frac{\mathrm{C}}{2}\right)[cos(\frac{\mathrm{A}+\mathrm{B}}{2}+co\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)]}$$$$=\frac{4cos\left(\frac{\mathrm{C}}{2}\right)cos\left(\frac{\mathrm{A}}{2}\right)sin\left(\frac{\mathrm{B}}{2}\right)}{4sin\left(\frac{\mathrm{C}}{2}\right)cos\left(\frac{\mathrm{A}}{2}\right)cos\left(\frac{\mathrm{B}}{2}\right)}=cot\left(\frac{\mathrm{C}}{2}\right)tan\left(\frac{\mathrm{B}}{2}\right)...▴$$$$.....\mathcal{M}.\mathcal{N}....$$

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