If \(\sin x+\sin y=\frac{\sqrt{3}+1}{2}\) and \(\cos x+\cos y=\) \(\frac{\sqrt{3}-1}{2}\), then \(\tan…
If \(\sin x+\sin y=\frac{\sqrt{3}+1}{2}\) and \(\cos x+\cos y=\) \(\frac{\sqrt{3}-1}{2}\), then \(\tan ^2\left(\frac{x-y}{2}\right)+\tan ^2\left(\frac{x+y}{2}\right)=\)
\(8+4 \sqrt{3}\)
\(6+4 \sqrt{3}\)
\(3+\sqrt{3}\)
\(12+6 \sqrt{3}\)
Solution
It is given that
\(\begin{array}{ll}
\sin x+\sin y & =\frac{\sqrt{3}+1}{2} \\
2 & 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{\sqrt{3}^2+1}{2}
\end{array}\)
\(\Rightarrow \quad \sin \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{\sqrt{3}+1}{4}\) ...(i)
and \(\quad \cos x+\cos y=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow \quad \cos \frac{x+y}{2} \cos \frac{x-y}{2}=\frac{\sqrt{3}-1}{4}\) ...(ii)
On dividing relation (i) and (ii), we get
\(\begin{aligned}
& \tan \frac{x+y}{2}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \\
\Rightarrow \quad & \tan ^2 \frac{x+y}{2}=\frac{3+1+2 \sqrt{3}}{3+1-2 \sqrt{3}}=\frac{2+\sqrt{3}}{2-\sqrt{3}} \\
\Rightarrow \quad & \quad \tan ^2 \frac{(x+y)}{2}=\frac{4+3+4 \sqrt{3}}{4-3}=7+4 \sqrt{3} \quad \ldots (iii)
\end{aligned}\)
On squaring and adding Eqs. (i) and (ii), we get
\(\begin{aligned}
& \cos ^2 \frac{x-y}{2}=\frac{1}{16} \times 2(3+1)=\frac{1}{2} \Rightarrow \sec ^2 \frac{x-y}{2}=2 \\
\Rightarrow & \tan ^2 \frac{x-y}{2}=\sec ^2 \frac{x-y}{2}-1=2-1=1
\end{aligned}\)
So, \(\tan ^2 \frac{x-y}{2}+\tan ^2 \frac{x+y}{2}=1+(7+4 \sqrt{3})\)
\(=8+4 \sqrt{3}\)
Hence, option (a) is correct.