If tan A - B -1- A - B -2-3- then the smallest positive value of B is-A- 25-24B- Iy -24 D- 15-24-

The correct option is B tan(A B)=1 A B=(2n+1)π/4 n=0 since question is the smallest value possible A B=π/4 A+B=2π π/6=11π/6 Subtract both the equations 2B=1 ...

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If tan A - B ...

Question

If $t a n (A - B) = 1, s e c (A + B) = \frac{2}{\sqrt{3}},$ then the smallest positive value of B is,

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Solution

The correct option is B
tan(A-B)=1

$A - B = (2 n + 1) \frac{π}{4}$

n=0 since question is the smallest value possible

$A - B = \frac{π}{4}$

$A + B = 2 π - \frac{π}{6} = 11 \frac{π}{6}$

Subtract both the equations

$2 B = 11 \frac{π}{6} - \frac{π}{4}$

$2 B = (22 - 3) \frac{π}{12}$

$B = 19 \frac{π}{24}$

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