Question

Without using trigonometric table, evaluate :

$\frac{2}{3}cose{c}^2{58}^o-\frac{2}{3}cot{58}^{o}tan{32}^o-\frac{5}{3}tan{13}^{o}tan{37}^{o}tan{45}^{o}tan{53}^{o}tan{77}^o$

Answer 1

$\frac{2}{3}cose{c}^2{58}^o-\frac{2}{3}cot{58}^{o}tan{32}^o-\frac{5}{3}tan{13}^{o}tan{37}^{o}tan{45}^{o}tan{53}^{o}tan{77}^o$

$=\frac{2}{3}cose{c}^2{58}^o-\frac{2}{3}cot{58}^{o}tan({90}^o-{58}^o)-\frac{5}{3}tan{13}^{o}tan{37}^{o}tan{45}^{o}tan({90}^o-{37}^o)tan({90}^o-{13}^o)$

$=\frac{2}{3}cose{c}^2{58}^o-\frac{2}{3}co{t}^2{58}^o-\frac{5}{3}tan{13}^{o}tan{37}^{o}tan{45}^{o}cot{37}^{o}cot{13}^o$

$=\frac{2}{3}(cose{c}^2{58}^o-co{t}^2{58}^o)-\frac{5}{3}tan{13}^{o}tan{37}^o\times 1\times \frac{1}{tan{37}^o}\times \frac{1}{tan{13}^o}$

$=\frac{2}{3}\times 1-\frac{5}{3}=-1$