Question

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

• A. $\frac{25}{24}\mathrm{\pi }$
• B. $\frac{19}{24}\mathrm{\pi }$
• C. $\frac{13}{24}\mathrm{\pi }$
• D. $\frac{11}{24}\mathrm{\pi }$

Answer 1

The correct option is B

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

Explanation for the correct option

Given that $\tan \left(A-B\right)=1$ and $\sec \left(A+B\right)=\frac{2}{\sqrt{3}}$

$\Rightarrow A-B={\tan }^{-1}\left(1\right)=\frac{\mathrm{\pi }}{4}$ and $A+B={\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\mathrm{\pi }}{6}$

To get positive value of $B$, $A-B$ should be less than $A+B$

$\therefore A+B$ can be re-written as $2\mathrm{\pi }-\frac{\mathrm{\pi }}{6}$

$\begin{array}{cccc}\therefore & \left(A+B\right)-\left(A-B\right)& =& 2\mathrm{\pi }-\frac{\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{4}\\ \Rightarrow & 2B& =& \frac{24\mathrm{\pi }-2\mathrm{\pi }-3\mathrm{\pi }}{12}\\ \Rightarrow & B& =& \frac{19}{24}\mathrm{\pi }\end{array}$

Hence the correct option is option(B) i.e. $\frac{19}{24}\mathrm{\pi }$