Question

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

• A. $\frac{25\pi }{24}$
• B. $\frac{19\pi }{24}$
• C. $\frac{13\pi }{24}$
• D. $\frac{7\pi }{24}$

Answer 1

Answer: (D)

$tan(A-B)=1$
$\Rightarrow A-B=45^{\circ }$ or $225^{\circ }$
$sec\left(A+B\right)=\frac{2}{\sqrt{3}}$
$\Rightarrow A+B=30^{\circ }$ or $330^{\circ }$
$A+B=330^{\circ }=2\pi -\frac{\pi }{6}=\frac{11\pi }{6}\dots \left(i\right)$
and $A-B=225^{\circ }=\frac{5\pi }{4}\dots \left(ii\right)$
Solving $(i)$ and $(ii)$, we get
$2B=\frac{11\pi }{6}-\frac{5\pi }{4}$
$\Rightarrow 2B=\frac{7\pi }{12}$
$\Rightarrow B=\frac{7\pi }{24}$