If t a n ( A − B ) = 1 , s e c ( A + B ) = 2 √ 3 , then the smallest positive value of B is, Options: A . &nbsp 25 24 π B . &nbsp 19 24 π C . &nbsp 13 24 π D . &nbsp 11 24 π

Answer: Option B : B tan(A-B)=1 A − B = ( 2 n + 1 ) π 4 n=0 since question is the smallest value possible A − B = π 4 A + B = 2 π − π 6 = 11 π 6 Subtract both the equations 2 B = 11 π 6 − π 4 2 B = ( 22 − 3 ) π 12 B = 19 π 24

If tan(A−B)=1 , sec(A+B)=2√3, then the sma

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,

Options:

A .

\frac{25}{24} π

B .

\frac{19}{24} π

C .

\frac{13}{24} π

D .

\frac{11}{24} π

Answer: Option B
:
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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