The correct option is C 92
sec254o−cot236ocosec257o−tan233o+2sin238osec252o−sin245o+2√3tan17otan60otan73o
⇒sec2(90o−36o)−cot236ocosec2(90o−33o)−tan233o+2sin238osec2(90o−38o)−sin245o+2√3tan(90o−73o)tan73otan60o
⇒cosec236o−cot236osec233o−tan233o+2sin238ocosec238o−(1√2)2+2√3cot73otan73o×√3
⇒11+2sin238o×1sin238o−12+2√3×1tan73o×tan73o×√3
[∵cosec20−cot20=1,sec20−tan20=1]
⇒1+2−12+2=5−12
⇒92