Question

Solve: $\frac{\cot {81}^o\cot {69}^o-\tan {225}^o}{\cot {261}^o+\tan {21}^o}$

Answer 1

$\frac{cot{81}^{o}cot{69}^o-tan{225}^o}{cot{261}^o+tan{21}^o}$

$=\frac{cot(90-9)cot(90-21)-tan(270-45)}{cot(270-9)+tan{21}^o}$

$=\frac{tan{9}^{0}tan{21}^0-cot{45}^o}{tan{9}^0+tan{21}^0}$

$=\frac{tan{9}^{0}tan{21}^0-1}{tan{9}^0+tan{21}^0}$

$=-\left[\frac{1-tan{9}^0+tan{21}^0}{tan{9}^0+tan{21}^0}\right]$

$=\left[{\left(\frac{tan{9}^0+tan{21}^0}{1-tan{9}^{0}tan{21}^0}\right)}^{-1}\right]$

$=-\left[\right(tan(9+21){)}^{-1}]$

$=-\frac{1}{tan{30}^o}$

$=-\frac{1}{\frac{1}{\sqrt{3}}}$

$=-\sqrt{3}$

$\Rightarrow \frac{cot{81}^{o}cot{69}^o-tan{225}^o}{cot{261}^o+tan{21}^o}=-\sqrt{3}$