$\frac{s i n 5 A c o s 2 A - s i n 6 A c o s A}{s i n A s i n 2 A - c o s 2 A c o s 3 A}$

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Solution

$\frac{s i n 5 A c o s 2 A - s i n 6 A c o s A}{s i n A s i n 2 A - c o s 2 A c o s 3 A}$

multiplying and dividing by 2

$\frac{2 s i n 5 A c o s 2 A - 2 s i n 6 A c o s A}{2 s i n A s i n 2 A - 2 c o s 2 A c o s 3 A}$

using
$2 s i n A c o s B = s i n (A + B) + s i n (A - B)$
$2 s i n A s i n B = c o s (A - B) - c o s (A + B)$
$2 c o s A c o s B = c o s (A + B) + c o s (A - B)$

$\frac{[s i n (5 A + 2 A) + s i n (5 A - 2 A)] - [s i n (6 A + A) + s i n (6 A - A)]}{[c o s (A - 2 A) - c o s (A + 2 A)] - [c o s (2 A + 3 A) + c o s (2 A - 3 A)]}$

$= \frac{s i n 7 A + s i n 3 A - s i n 7 A - s i n 5 A}{c o s A - c o s 3 A - c o s 5 A - c o s A}$

$= \frac{s i n 3 A - s i n 5 A}{- c o s 3 A - c o s 5 A} = \frac{s i n 3 A - s i n 5 A}{- [c o s 3 A + c o s 5 A]}$

using
$s i n A - s i n B = 2 s i n (\frac{A - B}{2}) c o s (\frac{A + B}{2})$
$c o s C + c o s D = 2 c o s (\frac{C + D}{2}) c o s (\frac{C - D}{2})$

So, $\frac{s i n 3 A - s i n 5 A}{- [c o s 3 A + c o s 5 A]}$
$= \frac{2 s i n (\frac{3 A - 5 A}{2}) c o s (\frac{3 A + 5 A}{2})}{- 2 c o s (\frac{3 A + 5 A}{2}) c o s (\frac{3 A - 5 A}{2})}$
$= \frac{- 2 s i n A c o s 4 A}{2 c o s 4 A c o s A}$ $[∵ s i n (- x) = - s i n x$ and $c o s (- x) = c o s x]$
$= t a n A$

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