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Question

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


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