If (A-B)=1, (A+B)=2 3 , then the smallest positive value of B is: - Vidyadip

MCQ

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

A
$\frac{25\pi}{24}$
✓
$\frac{19\pi}{24}$
C
$\frac{13\pi}{24}$
D
$\frac{11\pi}{24}$

✓

Answer

Correct option: B.

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

Given:
$\tan(\text{A - B})=1$ and $\sec(\text{A+B})=\frac{2}{\sqrt{3}}$
$\Rightarrow\text{A - B}=\frac{\pi}{4}\cdots(1)$ and $\text{A + B}=\frac{\pi}{4}\cdots(2)$
Adding these equations we get:
$2\text{A}=\frac{\pi}{4}+\frac\pi6$
$\Rightarrow\text{A}=\frac{5\pi}{24}$
$\Rightarrow$ Smallest possible value of $\text{B}=\pi-\frac{5\pi}{24}=\frac{19\pi}{24}.$

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