Question

2(cos 58 / sin 32) -- root of 3 (cos 38. cosec 52/tan 15 tan 60 tan 75)

Answer 1

Answer:

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Step-by-step explanation:

2(cos58/sin32)-√3(cos38*cosec52/tan15*tan60*tan75)

= 2{cos58/sin(90-58)}-√3[{cos38*cosec(90-38)}/tan(90-75)*tan60*tan75]

= 2{cos58/cos58}-√3[cos38*sec38/cot75*tan60*tan75]                  (1/tanФ = cotФ)

= 2*1-√3[1/tan60]                                               

= 2-√3(1/√3)

= 2-1

= 1