Without using trigonometric tables, evaluate the following:
2/3 cosec258° – 2/3 cot58°tan32° – 5/3 tan13°tan37°tan 45°tan 53°tan 77°
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2/3cosec258° – 2/3cot58°tan32° – 5/3 tan13°tan37°tan 45°tan 53°tan 77°
= 2/3(cosec258°- cot 58°tan32°) – 5/3 tan13°tan(90° – 13°) tan 37°tan(90° – 37°)(tan45°)
= 2/3[cosec258°- cot 58°tan(90°- 58°)] – 5/3 tan13°cot13°tan37°cot37°(1)
= 2/3(cosec258°- cot58°tan58°)- 5/3tan13°\(\frac{1}{tan13°}\)tan37°\(\frac{1}{tan37°}\)
= 2/3(cosec258°- cot258°- 5/3
= 2/3 – 5/3
= – 1
Hence proved