Question

Without using trigonometric tables, evaluate that :
$\frac{2}{3}{\text{cosec}}^2{58}^{\circ }-\frac{2}{3}\cot {58}^{\circ }\tan {32}^{\circ }-\frac{5}{3}\tan {13}^{\circ }\tan {37}^{\circ }\tan {45}^{\circ }\tan {53}^{\circ }\tan {77}^{\circ }$

Answer 1

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

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

$=\frac{2}{3}cose{c}^2{58}^{\circ }-\frac{2}{3}co{t}^2{58}^{\circ }-\frac{5}{3}tan{13}^{\circ }tan{37}^{\circ }tan{45}^{\circ }cot{37}^{\circ }cot{13}^{\circ }$ ........ $\tan (90-\theta )=\cot \theta$

$=\frac{2}{3}(cose{c}^2{58}^{\circ }-co{t}^2{58}^{\circ })-\frac{5}{3}tan{13}^{\circ }tan{37}^{\circ }\times 1\times \frac{1}{tan{37}^{\circ }}\times \frac{1}{tan{13}^{\circ }}$ ..... ${cosec}^{2}x=1+{\cot }^{2}x$

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