Question Number 189468 by cortano12 last updated on 17/Mar/23
$$\:\:\:\mathrm{If}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\mathrm{csc}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\:,\:\mathrm{then} \\ $$$$\:\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=? \\ $$
Commented by mehdee42 last updated on 17/Mar/23
$${the}\:{given}\:{relation}\:{is}\:{wrong}. \\ $$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)={cscx}−{cotx} \\ $$
Commented by cortano12 last updated on 17/Mar/23
$$\mathrm{why}\:\mathrm{wrong}? \\ $$
Commented by Frix last updated on 17/Mar/23
$$\mathrm{Why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{relation}? \\ $$
Commented by mehdee42 last updated on 17/Mar/23
$${sorry}.{it}\:{was}\:{my}\:{mistake}.{this}\:{is}\:{an}\:{equation}.\: \\ $$
Answered by horsebrand11 last updated on 17/Mar/23
$$\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)=\frac{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2tan}\:\left(\frac{{x}}{\mathrm{2}}\right)}−\frac{\mathrm{2tan}\:\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$\:{let}\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)={p}\Rightarrow{p}^{\mathrm{2}} =\mathrm{tan}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\: \\ $$$$\:\Rightarrow{p}=\frac{\mathrm{1}+{p}^{\mathrm{2}} }{\mathrm{2}{p}}−\frac{\mathrm{2}{p}}{\mathrm{1}+{p}^{\mathrm{2}} }\: \\ $$$$\Rightarrow{p}=\frac{\left(\mathrm{1}+{p}^{\mathrm{2}} \right)^{\mathrm{2}} −\left(\mathrm{2}{p}\right)^{\mathrm{2}} }{\mathrm{2}{p}\left(\mathrm{1}+{p}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow\mathrm{2}{p}^{\mathrm{2}} \left(\mathrm{1}+{p}^{\mathrm{2}} \right)={p}^{\mathrm{4}} −\mathrm{2}{p}^{\mathrm{2}} +\mathrm{1} \\ $$$$\:{let}\:{k}={p}^{\mathrm{2}} \:,\:{k}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{k}\left(\mathrm{1}+{k}\right)={k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{k}+\mathrm{2}{k}^{\mathrm{2}} ={k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1} \\ $$$$\Rightarrow{k}^{\mathrm{2}} +\mathrm{4}{k}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{k}={p}^{\mathrm{2}} =\frac{−\mathrm{4}+\sqrt{\mathrm{20}}}{\mathrm{2}}=\sqrt{\mathrm{5}}−\mathrm{2}\: \\ $$$$ \\ $$
Answered by Frix last updated on 17/Mar/23
$$\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:=\mathrm{csc}\:{x}\:−\mathrm{sin}\:{x} \\ $$$${t}=\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{t}}−\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${t}=\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} =−\mathrm{2}+\sqrt{\mathrm{5}}\:\left(\mathrm{for}\:{t}\in\mathbb{R}\right) \\ $$