Question

evaluate the following:
2(cos 58°/sin 32°)-√3(cos 38°×cosec 52°/ tan 15°×tan 60°×tan 75° )

Answer 1

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

Answer 2

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