Question

Find $\tan {65}^{\circ }-\tan {25}^{\circ }+2\tan {40}^{\circ }$

Answer 1

Let $\tan {65}^o$ be $x$ $\therefore \tan 25=\frac{1}{x}$

$\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x+\tan y}$

$x={65}^o;y={25}^o$

$x-y={40}^o$ $\tan {40}^o=\frac{\tan {65}^o-\tan {25}^o}{1+\tan {65}^o\tan {25}^o}$

Required expressions is

$\tan {40}^o(1+\tan {65}^o\tan {25}^o)+2\tan {40}^o$

Answer: $3\tan {40}^o+\tan {40}^o\tan {65}^o\tan {25}^o$

(which is positive quantity)

$\left(\tan {65}^o\right)\left(\tan {25}^o\right)=1$

$\therefore Ans=4\tan {40}^o$