Question

If $\tan (A-B)=1$ and $\sec (A+B)=\frac{2}{\sqrt{3}}$ then the smallest positive $A=$_______ and the smallest positive $B=$______.

• A. $-\frac{25\pi }{24}$ and $\frac{19\pi }{24}$
• B. $\frac{25\pi }{24}$ and $\frac{19\pi }{24}$
• C. $\frac{45\pi }{24}$ and $\frac{19\pi }{24}$
• D. $\frac{15\pi }{24}$ and $\frac{19\pi }{24}$

Answer 1

The correct option is B $\frac{25\pi }{24}$ and $\frac{19\pi }{24}$

$A-B=\frac{\pi }{4}$

$A+B=\frac{11\pi }{6}$

$2A=\frac{\pi }{4}+\frac{11\pi }{6}=\frac{50\pi }{24}$

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

$B=A-\frac{\pi }{4}=\frac{25\pi }{24}-\frac{\pi \times 6}{4\times 6}$

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