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Question Number 43877 by peter frank last updated on 16/Sep/18

tan (x/2)=cosec −sin x prove that tan^2 (x/2)=(-2±(√5))

$$\mathrm{tan}\:\frac{{x}}{\mathrm{2}}=\mathrm{cosec}\:−\mathrm{sin}\:{x}\:{prove}\:{that} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}=\left(-\mathrm{2}\pm\sqrt{\mathrm{5}}\right) \\ $$

Answered by ajfour last updated on 16/Sep/18

tan (x/2)=(1/(sin x))−sin x let tan (x/2) = t ⇒ sin x = ((2t)/(1+t^2 )) ⇒ t=((1+t^2 )/(2t))−((2t)/(1+t^2 )) ⇒ (1/(2t))−(t/2) = ((2t)/(1+t^2 )) ⇒ 1−t^4 = 4t^2 or (t^2 +2)^2 = 5 t^2 = tan^2 (x/2) = −2+(√5) .

$$\mathrm{tan}\:\frac{{x}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{sin}\:{x}}−\mathrm{sin}\:{x} \\ $$$${let}\:\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:=\:{t}\:\:\:\:\Rightarrow\:\:\mathrm{sin}\:{x}\:=\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\:{t}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}{t}}−\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{\mathrm{2}{t}}−\frac{{t}}{\mathrm{2}}\:=\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\mathrm{1}−{t}^{\mathrm{4}} \:=\:\mathrm{4}{t}^{\mathrm{2}} \\ $$$${or}\:\:\:\:\:\:\left({t}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{5} \\ $$$$\:\:\:{t}^{\mathrm{2}} \:=\:\mathrm{tan}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\:=\:−\mathrm{2}+\sqrt{\mathrm{5}}\:. \\ $$

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