Question

Prove that:

$\frac{2}{3}{\mathrm{cosec}}^{2}58°-\frac{2}{3}\cot 58°\tan 32°-\frac{5}{3}\tan 13°\tan 37°\tan 45°\tan 53°\tan 77°=-1$

Answer 1

$$\begin{gathered} \begin{array}{l}\frac{2}{3}\cos {\text{ec}}^2{58}^0-\frac{2}{3}\cot {58}^0\tan {32}^0-\frac{5}{3}\tan {13}^0\tan {37}^0\tan {45}^0\tan {53}^0\tan {77}^0\\ \end{array} \\ \begin{array}{l}\text{=}\frac{2}{3}(\cos {\text{ec}}^2{58}^0-\cot {58}^0\tan {32}^0)-\frac{5}{3}\tan {13}^0\tan ({90}^0-{13}^0)\tan {37}^0\tan ({90}^0-{37}^0)\left(\tan {45}^0\right)\\ \end{array} \\ \begin{array}{l}\text{=}\frac{2}{3}\{\cos {\text{ec}}^2{58}^0-\cot {58}^0\tan ({90}^0-{58}^0)\}-\frac{5}{3}\tan {13}^0\cot {13}^0\tan {37}^0\cot {37}^0\left(1\right)\\ \end{array} \\ \begin{array}{l}\text{=}\frac{2}{3}(\cos {\text{ec}}^2{58}^0-\cot {58}^0\cot {58}^0)-\frac{5}{3}\tan {13}^0\frac{1}{\tan {13}^0}\tan {37}^0\frac{1}{\tan {37}^0}\\ \end{array} \\ \begin{array}{l}\text{=}\frac{2}{3}(\cos {\text{ec}}^2{58}^0-{\cot }^2{58}^0)-\frac{5}{3}\\ \end{array} \\ \begin{array}{l}\text{=}\frac{2}{3}-\frac{5}{3}\\ \end{array} \\ \text{=}-1 \end{gathered}$$

Hence Proved