Question

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

• A. $8+4 \sqrt{3}$
• B. $6+4 \sqrt{3}$
• C. $3+\sqrt{3}$
• D. $12+6 \sqrt{3}$

Answer 1

Answer: (A)

It is given that
$\sin \,x+\sin\, y=\frac{\sqrt{3}+1}{2}$
$\Rightarrow \, 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{\sqrt{3}+1}{2}$
$\Rightarrow \, \sin \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{\sqrt{3}+1}{4}$
and $ \cos \,x+\cos y=\frac{\sqrt{3}-1}{2}$
$\Rightarrow \, 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{\sqrt{3}-1}{2}$
$\Rightarrow \,d \cos \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{\sqrt{3}-1}{4}\,\,\,\,\,\dots(ii)$
On dividing relation (i) and (ii), we get
$ \tan \frac{x+y}{2}=\frac{\sqrt{3}+1}{\sqrt{3}-1} $
$\Rightarrow \, \tan ^{2} \frac{x+y}{2}=\frac{3+1+2 \sqrt{3}}{3+1-2 \sqrt{3}}=\frac{2+\sqrt{3}}{2-\sqrt{3}} $
$\Rightarrow \, \tan ^{2} \frac{(x+y)}{2}=\frac{4+3+4 \sqrt{3}}{4-3}=7+4 \sqrt{3} \ldots (iii)$
On squaring and adding Eqs. (i) and (ii), we get
$\cos ^{2}\, \frac{x-y}{2}=\frac{1}{16} \times 2(3+1)=\frac{1}{2} $
$\Rightarrow \, \sec ^{2} \frac{x-y}{2}=2 $
$ \Rightarrow \,\tan ^{2} \frac{x-y}{2}=\sec ^{2} \frac{x-y}{2}-1=2-1=1 $
So, $ \tan ^{2} \frac{x-y}{2}+\tan ^{2} \frac{x+y}{2} =1+(7+4 \sqrt{3}) $