Question

If cosec(A+B)=2/√3 and tan(A-B)=1/√3 &A and B are acute angles find A and B
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Answer 1

$\csc(a + b) = \frac{2}{ \sqrt{3} } \\ = > \frac{1}{ \sin(a + b) } = \frac{2}{ \sqrt{3} } \\ = > \sin(a + b) = \frac{ \sqrt{3} }{2} \\ = > a + b = 60$

$tan(a - b) = \frac{1}{ \sqrt{3} } \\ = > a - b = 30$

a + b = 60° ----------(1)

a - b = 30° -----------(2)

From Equation (1) and (2),

a = 45° and b = 15° .

Both a and b angles are actute angle .

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Answer 2

Here is your answer mate...

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