Hellom shayanbaig_3!
I'll take a guess as what you meant . . .
Evaluate: .$\displaystyle \tan12^o - \cot78^o$ Note that 12° and 78° are complementary angles.
That is, they are in the same right triangle. Code:
*
/|
/ |
/ |
/12°|
/ |y
/ |
/ |
/ |
/ 78° |
* - - - - *
x
We see that: .$\displaystyle \begin{array}{ccccc}\tan12^o &=& \dfrac{opp}{adj} &=& \dfrac{x}{y} \\ \\[-3mm] \cot 78^o &=& \dfrac{adj}{opp} &=&\dfrac{x}{y} \end{array}$ . . . . They are equal !
Therefore: .$\displaystyle \tan12^o - \cot78^o \;=\;0$