If cosA+sinB=m and sinA+cosB=n, prove that $2 s i n (A + B) = m^{2} + n^{2} - 2$ .

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Solution

We have,
cosA+sinB=m and sinA+sinB=n
Now, $m^{2} + n^{2} - 2$
$= (c o s A + s i n B)^{2} = (s i n A + c o s B)^{2} - 2$
$= c o s^{2} A + s i n^{2} B + 2 c o s A s i n B + s i n^{2} A + c o s^{2} B + 2 s i n A c o s B - 2$
$= (s i n^{2} A + c o s^{2} A) + (s i n^{2} B + c o s^{2} B) + 2 c o s A s i n B + 2 s i n A c o s B - 2$
$= 1 + 1 + 2 c o s A s i n B + 2 s i n A c o s B - 2$
$= 2 + 2 (s i n A c o s B + c o s A s i n B) - 2$
$= 2 (s i n A c o s B + c o s A s i n B)$
$= 2 s i n (A + B)$
$∴ 2 s i n (A + B) = m^{2} + n^{2} - 2$
Hence proved.

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