Show that tg40 +tg20 = square root of3 sec10^2

Saturday, May 31, 2014

You need to remember the formula $\displaystyle{\tan{{a}}}+{\tan{{b}}}=$
sin(a+b)/(cos a*cos b)

$\displaystyle{\tan{{20}}}+{\tan{{40}}}=$
sin(20+40)/(cos 20*cos 40) = sin 60/(cos 20*cos 40)

To
calculate the product found at denominator, you need the help of
formula:

$\displaystyle{2}{\cos{{a}}}\cdot{\cos{{b}}}={\cos{{\left({a}+{b}\right)}}}+{\cos{{\left({a}-{b}\right)}}}$

$\displaystyle{2}{\cos{{20}}}\cdot{\cos{{40}}}={\cos{{\left({20}+{40}\right)}}}+$
cos(20-40)

$\displaystyle{2}{\cos{{20}}}\cdot{\cos{{40}}}={\cos{{60}}}+$
cos(-20)

The function cosine is even => cos(-20) =
cos 20

$\displaystyle{2}{\cos{{20}}}\cdot{\cos{{40}}}=\frac{{1}}{{2}}+$
cos(20)

$\displaystyle{\tan{{20}}}+{\tan{{40}}}={2}\frac{{\frac{{\sqrt{{3}}}}{{2}}}}{{\frac{{1}}{{2}}+}}$
cos(20))

$\displaystyle{\tan{{20}}}+{\tan{{40}}}={2}\frac{{{\left(\sqrt{{3}}\right)}}}{{{1}+}}$
2cos(20))

$\displaystyle{\cos{{20}}}={\cos{{2}}}\cdot{10}={2}{{\cos}}^{{2}}{10}-$
1

$\displaystyle{\tan{{20}}}+{\tan{{40}}}={2}\frac{{{\left(\sqrt{{3}}\right)}}}{{{1}+{2}{{\cos}}^{{2}}{10}-}}$
1)

Remove opposite
terms=>

=> $\displaystyle{\tan{{20}}}+{\tan{{40}}}=$
2((sqrt3))/(2cos^2 10) = (sqrt3)/(cos^2 10)

Remember that
$\displaystyle{\sec{{10}}}=\frac{{1}}{{{\cos{{10}}}}}$

$\displaystyle{\tan{{20}}}+{\tan{{40}}}={\left(\sqrt{{3}}\right)}\cdot{{\sec}}^{{2}}$
(10)

ANSWER: $\displaystyle{\tan{{20}}}+{\tan{{40}}}={\left(\sqrt{{3}}\right)}\cdot{{\sec}}^{{2}}$
(10)

Help Notes