  Question

If $$\tan\dfrac {\theta}{2}=\mathrm{cosec}\theta-\sin\theta$$, then

A
sin2θ2=2sin218o  B
cos2θ+2cosθ+1=0  C
sin2θ2=4sin218o  D
cos2θ+2cosθ1=0  Solution

The correct options are A $$\sin^2\dfrac {\theta}{2}=2 \sin^2 18^o$$ D $$\cos 2\theta+2 \cos\theta-1=0$$Simplifying, we get  $$\dfrac{\sin \dfrac {\theta}{2}}{\cos \dfrac {\theta}{2}}=\dfrac{1-\sin ^{2}\theta}{\sin \theta}$$ $$\dfrac{\sin \dfrac {\theta}{2}}{\cos \dfrac {\theta}{2}}=\dfrac{\cos ^{2}(\theta)}{2\sin \dfrac {\theta}{2}.\cos \dfrac {\theta}{2}}$$ $$2\sin ^{2}\dfrac {\theta}{2}=\cos ^{2}\theta$$ $$2\sin ^{2}\dfrac {\theta}{2}=(1-2\sin ^{2}\dfrac {\theta}{2})^{2}$$ Let, $$2\sin ^{2}\dfrac {\theta}{2}=t$$$$\implies t=(1-t)^{2}$$ $$\implies t=t^{2}-2t+1$$ $$\implies t^{2}-3t+1=0$$ $$\implies t=\dfrac{3\pm\sqrt{9-4}}{2}$$ $$\implies t=\dfrac{3\pm\sqrt{5}}{2}$$ Now $$|\sin \theta|\leq 1$$ $$\implies t=\dfrac{3-\sqrt{5}}{2}$$ Or  $$\sin ^{2}\dfrac {\theta}{2}=\dfrac{3-\sqrt{5}}{4}$$ $$\implies 2\left(\dfrac{3-\sqrt{5}}{8}\right)=2\sin ^{2}(18^{0})$$ Hence, option A is correct. $$\cos ^{2}\dfrac {\theta}{2}=1-\sin ^{2}\dfrac {\theta}{2}=\dfrac {1+\sqrt{5}}{4}$$  Option D  $$\cos 2\theta +2\cos \theta -1=0$$ $$-2{ \sin }^{ 2 }\theta +2\cos \theta =0$$ $$=-2+2{ \cos }^{ 2 }\theta +2\cos \theta =0$$ $$=-1+{ \cos }^{ 2 }\theta +\cos \theta =0$$ $${ \cos }^{ 2 }\theta =1-\cos \theta =2{ \sin }^{ 2 }\cfrac { \theta }{ 2 }$$ which is true (as proven in the first part)Hence, option A and D are correct.Maths

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