The smallest positive values of x and y which | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $	an(x - y) = 1$,we have $x - y = \frac{\pi}{4}, \frac{5\pi}{4}, \dots$ (considering positive values).Given $\sec(x + y) = \frac{2}{\sqrt{3

Vedclass · Accepted Answer

$a$ or $b$ both

Vedclass · Answer

(D) Given $	an(x - y) = 1$,we have $x - y = \frac{\pi}{4}, \frac{5\pi}{4}, \dots$ (considering positive values).Given $\sec(x + y) = \frac{2}{\sqrt{3}}$,we have $\cos(x + y) = \frac{\sqrt{3}}{2}$,so $x + y = \frac{\pi}{6}, \frac{11\pi}{6}, \dots$ (considering positive values).Since $x, y > 0$,we require $x + y > x - y$.Case $1$: $x + y = \frac{11\pi}{6}$ and $x - y = \frac{\pi}{4}$. Adding gives $2x = \frac{22\pi + 3\pi}{12} = \frac{25\pi}{12} \Rightarrow x = \frac{25\pi}{24}$. Then $y = \frac{11\pi}{6} - \frac{25\pi}{24} = \frac{44\pi - 25\pi}{24} = \frac{19\pi}{24}$.Case $2$: $x + y = \frac{11\pi}{6}$ and $x - y = \frac{5\pi}{4}$. Adding gives $2x = \frac{22\pi + 15\pi}{12} = \frac{37\pi}{12} \Rightarrow x = \frac{37\pi}{24}$. Then $y = \frac{11\pi}{6} - \frac{37\pi}{24} = \frac{44\pi - 37\pi}{24} = \frac{7\pi}{24}$.Both pairs are valid positive solutions.

The smallest positive values of $x$ and $y$ which satisfy $\tan (x - y) = 1$ and $\sec (x + y) = \frac{2}{\sqrt{3}}$ are

Similar Questions

The value of $\cos y \cos (\frac{\pi}{2} - x) - \cos (\frac{\pi}{2} - y) \cos x + \sin y \cos (\frac{\pi}{2} - x) + \cos x \sin (\frac{\pi}{2} - y)$ is zero,if

The number of solutions to the equation $\cos^4 x + \frac{1}{\cos^2 x} = \sin^4 x + \frac{1}{\sin^2 x}$ in the interval $[0, 2\pi]$ is

The number of solutions of the equation $\sin^{65}x - \cos^{65}x = -1$ for $x \in (-\pi, \pi)$ is:

The general solution of $a \cos x + b \sin x = c,$ where $a, b, c$ are constants is:

If $(2\cos x - 1)(3 + 2\cos x) = 0$ for $0 \le x \le 2\pi$,then $x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module