Question

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

• A. $\frac{25\pi }{24},\frac{19\pi }{24}$
• B. $\frac{19\pi }{24},\frac{25\pi }{24}$
• C. $\frac{31\pi }{24},\frac{13\pi }{24}$
• D. $\frac{13\pi }{24},\frac{31\pi }{24}$

Answer 1

The correct option is A $\frac{25\pi }{24},\frac{19\pi }{24}$

$\tan (A-B)=1$

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

$\tan 2A=\tan \left(\right(A+B)+(A-B\left)\right)=\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B)\tan (A-B)}$

$=\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}}=\frac{1+\sqrt{3}}{\sqrt{3}-1}=2+\sqrt{3}$

$2A=n\pi +\frac{5\pi }{12}\Rightarrow A=\frac{n\pi }{2}+\frac{5\pi }{24}$

$\tan 2B=\tan \left(\right(A+B)-(A-B\left)\right)=\frac{\tan (A+B)-\tan (A-B)}{1+\tan (A+B)\tan (A-B)}$

$=\frac{\frac{1}{\sqrt{3}}-1}{1+\frac{1}{\sqrt{3}}}=\frac{1-\sqrt{3}}{\sqrt{3}+1}=2-\sqrt{3}$

$2B=n\pi +\frac{\pi }{12}\Rightarrow B=\frac{n\pi }{2}+\frac{\pi }{24}$

$A=\frac{19\pi }{24},B=\frac{25\pi }{24}$