Evaluate:
$tan 65^{o} = tan 25^{o} + 2 tan 40^{o}$ ?

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Solution

$t a n 65 - t a n 40 = t a n 25 + t a n 40.$
Using,
$t a n A - t a n B = \frac{s i n (A - B)}{c o s A c o s B}$ and

$t a n A + t a n B$ = $\frac{s i n (A + B)}{c o s A c o s B}$

we get to prove, $\frac{s i n 25}{c o s 65}$

$= \frac{s i n 65}{c o s 25 s i n 25}$

$= c o s (90 - 25)$

$= c o s 65 a n d s i n 65$

$= c o s (90 - 65)$

$= c o s 25$

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