Question #9b18b

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Question #9b18b

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Answer 1 · 2017-03-30T05:26:19

Let

$\frac{θ}{2} = 75^{\circ}$ $theta/2=75^@$

$\Rightarrow θ = 150^{\circ}$ $=>theta=150^@$

$\Rightarrow tan θ = {tan 150}^{\circ}$ $=>tantheta=tan150^@$

$\Rightarrow \frac{2 tan (\frac{θ}{2})}{1 - {tan}^{2} (\frac{θ}{2})} = {tan (180 - 30)}^{\circ}$ $=>(2tan(theta/2))/(1-tan^2(theta/2))=tan(180-30)^@$

putting $tan (\frac{θ}{2}) = x$ $tan(theta/2)=x$

$\Rightarrow \frac{2 x}{1 - x^{2}} = - {tan 30}^{\circ} = - \frac{1}{\sqrt{3}}$ $=>(2x)/(1-x^2)=-tan30^@=-1/sqrt3$

$\Rightarrow x^{2} - 2 \sqrt{3} x - 1 = 0$ $=>x^2-2sqrt3x-1=0$

$\Rightarrow x = \frac{2 \sqrt{3} + \sqrt{{(2 \sqrt{3})}^{2} - 4 \cdot 1 \cdot (- 1)}}{2}$ $=>x=(2sqrt3+sqrt((2sqrt3)^2-4*1*(-1)))/2$
[Negative root of x neglected as $tan 75 > 0$ $tan75>0$ ]

$\Rightarrow tan (\frac{θ}{2}) = \frac{2 \sqrt{3} + \sqrt{16}}{2}$ $=>tan(theta/2)=(2sqrt3+sqrt16)/2$

$\Rightarrow tan 75 = \frac{2 \sqrt{3} + 4}{2}$ $=>tan75=(2sqrt3+4)/2$

$\Rightarrow tan 75 = 2 + \sqrt{3}$ $=>tan75=2+sqrt3$

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Question #9b18b

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