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Question Number 189468 by cortano12 last updated on 17/Mar/23

If tan ((x/2))= csc x−sin x , then tan^2 ((x/2))=?

$$\:\:\:\mathrm{If}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\mathrm{csc}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\:,\:\mathrm{then} \\ $$$$\:\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=? \\ $$

Commented by mehdee42 last updated on 17/Mar/23

the given relation is wrong. tan((x/2))=cscx−cotx

$${the}\:{given}\:{relation}\:{is}\:{wrong}. \\ $$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)={cscx}−{cotx} \\ $$

Commented by cortano12 last updated on 17/Mar/23

why wrong?

$$\mathrm{why}\:\mathrm{wrong}? \\ $$

Commented by Frix last updated on 17/Mar/23

Why do you think it′s a relation?

$$\mathrm{Why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{relation}? \\ $$

Commented by mehdee42 last updated on 17/Mar/23

sorry.it was my mistake.this is an equation.

$${sorry}.{it}\:{was}\:{my}\:{mistake}.{this}\:{is}\:{an}\:{equation}.\: \\ $$

Answered by horsebrand11 last updated on 17/Mar/23

tan ((x/2))=((1+tan^2 ((x/2)))/(2tan ((x/2))))−((2tan ((x/2)))/(1+tan^2 ((x/2)))) let tan ((x/2))=p⇒p^2 =tan^2 ((x/2)) ⇒p=((1+p^2 )/(2p))−((2p)/(1+p^2 )) ⇒p=(((1+p^2 )^2 −(2p)^2 )/(2p(1+p^2 ))) ⇒2p^2 (1+p^2 )=p^4 −2p^2 +1 let k=p^2 , k>0 ⇒2k(1+k)=k^2 −2k+1 ⇒2k+2k^2 =k^2 −2k+1 ⇒k^2 +4k−1=0 ⇒k=p^2 =((−4+(√(20)))/2)=(√5)−2

$$\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)=\frac{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2tan}\:\left(\frac{{x}}{\mathrm{2}}\right)}−\frac{\mathrm{2tan}\:\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$\:{let}\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)={p}\Rightarrow{p}^{\mathrm{2}} =\mathrm{tan}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\: \\ $$$$\:\Rightarrow{p}=\frac{\mathrm{1}+{p}^{\mathrm{2}} }{\mathrm{2}{p}}−\frac{\mathrm{2}{p}}{\mathrm{1}+{p}^{\mathrm{2}} }\: \\ $$$$\Rightarrow{p}=\frac{\left(\mathrm{1}+{p}^{\mathrm{2}} \right)^{\mathrm{2}} −\left(\mathrm{2}{p}\right)^{\mathrm{2}} }{\mathrm{2}{p}\left(\mathrm{1}+{p}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow\mathrm{2}{p}^{\mathrm{2}} \left(\mathrm{1}+{p}^{\mathrm{2}} \right)={p}^{\mathrm{4}} −\mathrm{2}{p}^{\mathrm{2}} +\mathrm{1} \\ $$$$\:{let}\:{k}={p}^{\mathrm{2}} \:,\:{k}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{k}\left(\mathrm{1}+{k}\right)={k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{k}+\mathrm{2}{k}^{\mathrm{2}} ={k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{1} \\ $$$$\Rightarrow{k}^{\mathrm{2}} +\mathrm{4}{k}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{k}={p}^{\mathrm{2}} =\frac{−\mathrm{4}+\sqrt{\mathrm{20}}}{\mathrm{2}}=\sqrt{\mathrm{5}}−\mathrm{2}\: \\ $$$$ \\ $$

Answered by Frix last updated on 17/Mar/23

tan (x/2) =csc x −sin x t=((t^2 +1)/(2t))−((2t)/(t^2 +1)) t=(((t^2 −1)^2 )/(2t(t^2 +1))) t^4 +4t^2 −1=0 t^2 =−2+(√5) (for t∈R)

$$\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:=\mathrm{csc}\:{x}\:−\mathrm{sin}\:{x} \\ $$$${t}=\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{t}}−\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${t}=\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} =−\mathrm{2}+\sqrt{\mathrm{5}}\:\left(\mathrm{for}\:{t}\in\mathbb{R}\right) \\ $$

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