Dayanara Terry

2022-07-03

Prove $\frac{2\mathrm{sec}\theta +3\mathrm{tan}\theta +5\mathrm{sin}\theta -7\mathrm{cos}\theta +5}{2\mathrm{tan}\theta +3\mathrm{sec}\theta +5\mathrm{cos}\theta +7\mathrm{sin}\theta +8}=\frac{1-\mathrm{cos}\theta }{\mathrm{sin}\theta }$

isscacabby17

Expert

Let $\mathrm{cos}\theta =c,\mathrm{sin}\theta =s$. Then,
$\begin{array}{rl}& s\left(2+3s+5sc-7{c}^{2}+5c\right)-\left(1-c\right)\left(2s+3+5{c}^{2}+7cs+8c\right)\\ & =5{c}^{3}+5c{s}^{2}-5c+3{c}^{2}+3{s}^{2}-3\\ & =5c\left({c}^{2}+{s}^{2}-1\right)+3\left({c}^{2}+{s}^{2}-1\right)\\ & =0\end{array}$
from which the claim follows.

Joel French

Expert

hint $\frac{1-\mathrm{cos}\left(\theta \right)}{sin\left(\theta \right)}=\mathrm{tan}\left(\theta /2\right)$ so now convert everything to tan by using $\mathrm{sin}\left(2x\right)=\frac{2tan\left(x\right)}{1+{\mathrm{tan}}^{2}\left(x\right)},\mathrm{cos}\left(2x\right)=\frac{1-{\mathrm{tan}}^{2}\left(x\right)}{1+{\mathrm{tan}}^{2}\left(x\right)}$ then just simplify it. as $1+{\mathrm{tan}}^{2}\left(x\right)$ cancels off from the expression by taking lcm then use $tan\left(x\right)=t$ for algebraic simplifications. I dont think there's a nice short way to do it.