Answers

2014-08-13T20:41:40+05:30
Tan15=tan(60-45)=tan60-tan45/1+tan60*tan45=√3-1/√3+1
tan75=tan(45+30)=tan45+tan30/1-tan45*tan60=√3+1/√3-1
sin²52=0.62
=2(cos58/sin32)-√3(cos38*sin52/tan15*tan75*tan60)
=2(cos(90-32)/sin32)-√3(cos(90-52)*sin52/√3-1/√3+1*√3+1/√3-1*√3)
=2(sin32/sin32)-√3(sin52*sin52/√3)
=2(1)-√3(sin²52)
=2-√3(0.62/√3)
=2-(0.62)
=2-0.62
=1.32
2 4 2
The Brainliest Answer!
2014-08-14T01:16:44+05:30
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
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