If sin x+sin y=3+12 and cos x+cos y=3-12, then tan2x-y2+tan2x+y2=

sin x+sin y=3+12 2sin x+y2 cosx-y2=3+12 cos x+cos y=3-12 2cos x+y2 cosx-y2=3+12 2sinx+y2cosx-y22coss+y2cosx-y2=3+123-12 tanx+y2=3+13-1×3+13+1 =4+232 =2+3(sin x+sin y)2+(cos x+cos y)2=3+122+3-122 sin2x+sin2y+2sin x sin y+cos2x+cos2y+2cos x cos y=2(3+1)4 2+2(cos x cos y+sin x sin y)=2 cos(x-y)=0 1-tan2x-y21+tan2x-y2=0 tan2x-y2=1 tan2x+y2+tan2x-y2 =4+3+23+1 =8+23

If sin x+sin y=3+12 and cos x+cos y=3-12, then tan2x-y2+tan2x+y2=

sin x+sin y=3+12 2sin x+y2 cosx-y2=3+12 cos x+cos y=3-12 2cos x+y2 cosx-y2=3+12 2sinx+y2cosx-y22coss+y2cosx-y2=3+123-12 tanx+y2=3+13-1×3+13+1 =4+232 =2+3(sin x+sin y)2+(cos x+cos y)2=3+122+3-122 sin2x+sin2y+2sin x sin y+cos2x+cos2y+2cos x cos y=2(3+1)4 2+2(cos x cos y+sin x sin y)=2 cos(x-y)=0 1-tan2x-y21+tan2x-y2=0 tan2x-y2=1 tan2x+y2+tan2x-y2 =4+3+23+1 =8+23

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If $\sin x + \sin y = \frac{\sqrt{3} + 1}{2}$ and $\cos x + \cos y = \frac{\sqrt{3} - 1}{2},$ then $\tan^{2} (\frac{x - y}{2}) + \tan^{2} (\frac{x + y}{2}) =$

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$8 + 4 \sqrt{3}$

$6 + 4 \sqrt{3}$

$3 + \sqrt{3}$

$12 + 6 \sqrt{3}$

answer is A.

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Detailed Solution

$\sin x + \sin y = \frac{\sqrt{3} + 1}{2} 2 \sin \frac{x + y}{2} \cos \frac{x - y}{2} = \frac{\sqrt{3} + 1}{2} \cos x + \cos y = \frac{\sqrt{3} - 1}{2} 2 \cos \frac{x + y}{2} \cos \frac{x - y}{2} = \frac{\sqrt{3} + 1}{2} \frac{2 \sin (\frac{x + y}{2}) \cos (\frac{x - y}{2})}{2 \cos (\frac{s + y}{2}) \cos (\frac{x - y}{2})} = \frac{\frac{\sqrt{3} + 1}{2}}{\frac{\sqrt{3} - 1}{2}} \tan (\frac{x + y}{2}) = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{4 + 2 \sqrt{3}}{2} = 2 + \sqrt{3}$

$\begin{array}{lcl} \begin{array}{rcl} {(\sin x + \sin y)}^{2} + {(\cos x + \cos y)}^{2} & = & {(\frac{\sqrt{3} + 1}{2})}^{2} + {(\frac{\sqrt{3} - 1}{2})}^{2} \end{array} \\ \sin^{2} x + \sin^{2} y + 2 \sin x \sin y + \cos^{2} x + \cos^{2} y + 2 \cos x \cos y = \frac{2 (3 + 1)}{4} \\ \begin{array}{rcl} 2 + 2 (\cos x \cos y + \sin x \sin y) & = & 2 \end{array} \\ \cos (x - y) = 0 \\ \frac{1 - \tan^{2} (\frac{x - y}{2})}{1 + \tan^{2} (\frac{x - y}{2})} \begin{array}{cl} = & 0 \end{array} \\ \tan^{2} (\frac{x - y}{2}) \begin{array}{cl} = & 1 \end{array} \\ \tan^{2} (\frac{x + y}{2}) + \tan^{2} (\frac{x - y}{2}) \\ \begin{array}{cl} = & 4 + 3 + 2 \sqrt{3} + 1 \end{array} \\ \begin{array}{cl} = & 8 + 2 \sqrt{3} \end{array} \end{array}$