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
$\tan \frac{x+y}{2}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$\Rightarrow {\tan }^2\frac{x+y}{2}=\frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$
$\Rightarrow {\tan }^2\frac{(x+y)}{2}=\frac{4+3+4\sqrt{3}}{4-3}=7+4\sqrt{3}\dots (iii)$
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, ${\tan }^2\frac{x-y}{2}+{\tan }^2\frac{x+y}{2}=1+(7+4\sqrt{3})$

=8+43​