Identity transformations of Trigonometric Expressions.
prove the following identities.
$\frac{s i n 2 x - s i n 3 x - s i n 4 x}{c o s 2 x - c o s 3 x + c o s 4 x} = t a n 3 x .$

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Solution

$\frac{s i n 2 x - s i n 3 x - s i n 4 x}{c o s 2 x - c o s 3 x + c o s 4 x} = t a n 3 x$
Let's take
LHS $= \frac{s i n 2 x - s i n 3 x - s i n 4 x}{c o s 2 x - c o s 3 x + c o s 4 x}$
from identify
$∴ s i n A - s i n B = 2 s i n (\frac{A + B}{2}) c o s (\frac{A - B}{2})$
and $c o s A + c o s B = 2 c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2})$
So, LHS $= \frac{(s i n 2 x - s i n 4 x) - s i n 3 x}{(c o s 2 x + c o s 4 x) - c o s 3 x}$
using above properties
LHS $= \frac{2 s i n (\frac{6 x}{2}) c o s (\frac{2 x - 4 x}{2}) - s i n 3 x}{2 c o s (\frac{2 x + 4 x}{2}) c o s (\frac{2 x - 4 x}{2}) - c o s 3 x}$
LHS $= \frac{2 s i n 3 x c o s x - s i n 3 x}{2 c o s 3 x c o s x - c o s 3 x}$ } $c o s (- x) = c o s x$
Lets take $s i n 3 x$ and $c o s 3 x$ conman
LHS $= \frac{s i n 3 x}{c o s 3 x} (\frac{2 c o s x - 1}{2 c o s x - 1})$
$L H S = t a n 3 x = R H S$
Hence proved.

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