If tan (A - B) = 1 and sec (A + B) = frac{2}{ | vedclass

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

Question

(B) Given $	an (A - B) = 1$. Since $	an(\pi/4) = 1$,we have $A - B = \frac{\pi}{4} + n\pi$. For the smallest positive value,consider $A - B = \frac{

Vedclass · Accepted Answer

$\frac{19}{24}\pi$

Vedclass · Answer

(B) Given $	an (A - B) = 1$. Since $	an(\pi/4) = 1$,we have $A - B = \frac{\pi}{4} + n\pi$. For the smallest positive value,consider $A - B = \frac{\pi}{4}$ $(i)$.Given $\sec (A + B) = \frac{2}{\sqrt{3}}$. Since $\sec(\pi/6) = \frac{2}{\sqrt{3}}$,we have $A + B = 2n\pi \pm \frac{\pi}{6}$.To find the smallest positive $B$,we solve for $B$ using $B = \frac{(A+B) - (A-B)}{2}$.Using $A + B = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$ and $A - B = \frac{\pi}{4}$:$2B = \frac{11\pi}{6} - \frac{\pi}{4} = \frac{22\pi - 3\pi}{12} = \frac{19\pi}{12}$.Therefore,$B = \frac{19\pi}{24}$.

If $\tan (A - B) = 1$ and $\sec (A + B) = \frac{2}{\sqrt{3}}$,then the smallest positive value of $B$ is

Similar Questions

The minimum value of $(8 \sec^2 \theta + 2 \cos^2 \theta)$ is equal to :-

If $\sin \alpha = -\frac{3}{5},$ where $\pi < \alpha < \frac{3\pi}{2},$ then $\cos \frac{\alpha}{2} = $

The value of $\cos^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{4} + \cos^2 \frac{5\pi}{12}$ is

$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} = $

The least value of $E = \frac{25 \sec^4 x - 50 \sec^2 x + 74}{\tan^2 x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module