Question

Find the smallest positive values of x and y satisfying $x-y=\frac{\pi }{4}$ and $\cot x+\cot y=2$

• A. The least positive values of x and y are $\frac{\pi }{6}$ and$\frac{5\pi }{12}$ respectively
• B. The least positive values of x and y are $\frac{5\pi }{12}$ and $\frac{\pi }{6}$ respectively
• C. The least positive values of x and y are $\frac{5\pi }{6}$ and $\frac{\pi }{3}$ respectively
• D. The least positive values of x and y are $\frac{\pi }{3}$ and $\frac{5\pi }{6}$ respectively

Answer 1

The correct option is B The least positive values of x and y are $\frac{5\pi }{12}$ and $\frac{\pi }{6}$ respectively

Given,

$x-y=\frac{\pi }{4}$ (i)

$\cot x+\cot y=2$ (ii)

From (ii), $\sin (x+y)=2\sin x\sin y$
$=\cos (x-y)-\cos (x+y)$
$=\cos \frac{\pi }{4}-\cos (x+y)$

$\sin (x+y)+\cos (x+y)=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{2}}\sin (x+y)+\frac{1}{\sqrt{2}}\cos (x+y)=\frac{1}{2}$

$\cos (x+y-\frac{\pi }{4})=\cos \frac{\pi }{3}$

$\Rightarrow$ $x+y-\frac{\pi }{4}=2n\pi \pm \frac{\pi }{3}$, $n\in Z$

$x+y=2n\pi \pm \frac{\pi }{3}+\frac{\pi }{4}$ (iii)

for $n=0$, $x+y=\frac{7\pi }{12}$ $(\because x,y>0)$ (iv)

From (i) and (iv), $x=\frac{7\pi }{12}$, $y=\frac{\pi }{6}$

Hence, the least positive values of x and y are $\frac{5\pi }{12}$ and $\frac{\pi }{6}$ respectively