Without using trigonometric tables, evaluate the following:

2/3 cosec^{2}58° – 2/3 cot58°tan32° – 5/3 tan13°tan37°tan 45°tan 53°tan 77°

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2/3cosec

^{2}58° – 2/3cot58°tan32° – 5/3 tan13°tan37°tan 45°tan 53°tan 77°= 2/3(cosec

^{2}58°- cot 58°tan32°) – 5/3 tan13°tan(90° – 13°) tan 37°tan(90° – 37°)(tan45°)= 2/3[cosec

^{2}58°- cot 58°tan(90°- 58°)] – 5/3 tan13°cot13°tan37°cot37°(1)= 2/3(cosec

^{2}58°- cot58°tan58°)- 5/3tan13°\(\frac{1}{tan13°}\)tan37°\(\frac{1}{tan37°}\)= 2/3(cosec

^{2}58°- cot^{2}58°- 5/3= 2/3 – 5/3

= – 1

Hence proved