Question

Find the smallest positive value of $x$ and $y$ satisfying the equations : $x-y=\frac{\pi }{4}$ & $\cot x+\cot y=2$.

• A. $x=\frac{\pi }{12}$, $y=\frac{\pi }{3}$
• B. $x=\frac{5\pi }{12}$, $y=\frac{\pi }{6}$
• C. $x=\frac{7\pi }{12}$, $y=\frac{2\pi }{3}$
• D. $x=\frac{5\pi }{12}$, $y=\frac{\pi }{4}$

Answer 1

Answer: (B)

$x=\frac{5\pi }{12}$, $y=\frac{\pi }{6}$

Given, $x-y=\frac{\pi }{4}$ ...(i)
and $\cot x+\cot y=2$ ....(ii)
From(ii), $\sin (x+y)=2\sin x.\sin y$
$=\cos (x-y)-\cos (x+y)$
$=\cos \frac{\pi }{4}-\cos (x+y)$
$\Rightarrow \sin (x+y)+\cos (x+y)=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$
$\Rightarrow \frac{1}{\sqrt{2}}\sin (x+y)+\frac{1}{\sqrt{2}}\cos (x+y)=\frac{1}{2}$
$\Rightarrow \cos (x+y-\frac{\pi }{4})=\cos \frac{\pi }{3}$
$\Rightarrow x+y-\frac{\pi }{4}=2n\pi \pm \frac{\pi }{3}$
$\Rightarrow x+y=2n\pi \pm \frac{\pi }{3}+\pm \frac{\pi }{4}$ ...(iii)
from $n=0,x+y=\frac{7\pi }{12}$ (since $x,y>0$) ...(iv)
From (i) and (iv), $x=\frac{5\pi }{12},y=\frac{\pi }{6}$
Hence, least positive values of $x$ and $y$ are $\frac{5\pi }{12}$ and $\frac{\pi }{6}$ respectively.