In-ABC-prove-that-1-cosA-cosB-cosC-1-cosA-cosB-cosC-tan-B-2-cot-C-2-

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

In △ ABC,

prove that \frac{1 + cosA - cosB + cosC}{1 + cosA + cosB - cosC} = \tan \frac{B}{2} \cot \frac{C}{2}

Answered by mnjuly1970 last updated on 20/Aug/20

\frac{(1 + c o s (C)) + (c o s (A) - c o s (B))}{(1 - c o s (C)) + (c o s (A) + c o s (B))}

= \frac{2 {c o s}^{2} (\frac{C}{2}) - 2 s i n (\frac{A + B}{2}) s i n (\frac{A - B}{2})}{2 {s i n}^{2} (\frac{C}{2}) + 2 c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2})}

= \frac{2 c o s (\frac{C}{2}) [s i n (\frac{A + B}{2}) - s i n (\frac{A - B}{2})]}{2 s i n (\frac{C}{2}) [c o s (\frac{A + B}{2} + c o (\frac{A - B}{2})]}

= \frac{4 c o s (\frac{C}{2}) c o s (\frac{A}{2}) s i n (\frac{B}{2})}{4 s i n (\frac{C}{2}) c o s (\frac{A}{2}) c o s (\frac{B}{2})} = c o t (\frac{C}{2}) t a n (\frac{B}{2}) \dots ▴

\dots . . M . N \dots .