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Sin(A+B)Sin(A-B)

If tan A+B ...

Question

If $tan (A + B) = \sqrt{3}$ and $tan (A - B) = \frac{1}{\sqrt{3}}; 0^{o} < A + B \leq 90^{o}; A > B,$ find $A$ and $B$ .

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Solution

$tan (A + B) = \sqrt{3}$
$A + B = \frac{π}{3}$
$tan (A - B) = \frac{1}{\sqrt{3}}$
$A - B = \frac{π}{6}$
$∴ A = \frac{π}{2},$ $B = \frac{- π}{6}$

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Similar questions

tan (A + B) = \sqrt{3}

tan (A - B) = \frac{1}{\sqrt{3}}; 0^{o} < A + B \leq 90^{o}; A > B

, find A and B.

tan (A + B) = \sqrt{3}

tan (A - B) = \frac{1}{\sqrt{3}}; 0^{o} < A + B \leq 90^{o}; A > B

then the value of A and B

tan (A + B) = \frac{1}{\sqrt{3}}

tan (A - B) = \frac{1}{\sqrt{3}}

0^{o} < A + B \leq 90^{o}

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