Question

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

• A. $25\pi /24$
• B. $13\pi /24$
• C. $19\pi /24$
• D. $11\pi /24$

Answer 1

The correct option is A $25\pi /24$

$A-B=n\pi =\pi /4$ $\left[as\tan \right(A-B)=1].......\left(1\right)$

again $\frac{1}{\cos (A+B)}=\sec (A+B)=\frac{2}{\sqrt{3}}$

$\therefore \cos (A+B)=\frac{\sqrt{3}}{2}$

$\therefore A+B=2n\pi +\pi /6.......\left(2\right)$

$\therefore$ Solving $\left(1\right)$ and $\left(2\right)$

$A=2n\pi +\frac{1}{2}(\frac{\pi }{4}+\frac{\pi }{6})$

$B=2n\pi +\frac{1}{2}(\frac{\pi }{6}-\frac{\pi }{4})$

for $x=0$

smallest value of $B=\frac{1}{2}(\frac{\pi }{6}-\frac{\pi }{4})$

$=\frac{25\pi }{24}$