If tan A - B -1- A - B -2-3then the smallest positive value of B isA- 25 -24B- 19 -24C- 13 -24D- 11 -24

The correct option is B 19 π/24 Given: tan(A B)=1 and sec(A+B)= 2/√(3) ⇒ A B= π/4 ....(1) and A+B= π/6 ....(2) Adding these equations we get: 2A= π/4 ...

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If tan A - B ...

Question

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

\frac{25 π}{24}

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\frac{19 π}{24}

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\frac{13 π}{24}

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\frac{11 π}{24}

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Solution

The correct option is B $\frac{19 π}{24}$

Given:
tan(A-B)=1 and sec(A+B)= $\frac{2}{\sqrt{3}}$
$\Rightarrow A - B = \frac{π}{4}$ ....(1) and
$A + B = \frac{π}{6}$ ....(2)
Adding these equations we get:
$2 A = \frac{π}{4} + \frac{π}{6}$
$\Rightarrow A = \frac{5 π}{24}$
$\Rightarrow$ Smallest possible value of B
$= π - \frac{5 π}{24} = \frac{19 π}{24}$

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If tan(A-B)=1, sec(A+B)=2√3, then the smallest positive value of B is

The correct option is B 19π24 Given: tan(A-B)=1 and sec(A+B)= 2√3 ⇒A−B=π4 ....(1) and A+B=π6 ....(2) Adding these equations we get: 2A=π4+π6 ⇒A=5π24 ⇒ Smallest possible value of B =π−5π24=19π24

If tan(A-B)=1, $s e c (A + B) = \frac{2}{\sqrt{3}},$
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