1.

Cos58/sin32-root3cos38cosec52/tan15tan60 tan75

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\frac{ \cos(58) }{ \sin(90 - 58) } - \frac{ \sqrt{3 } \cos(38) \ \csc (90 - 38) }{ \<klux>TAN</klux>(90 - 75) \tan(60) \tan(75) }
We know that Sin(90-x)=COSX
Cosec(90-x) = Secx
Tan (90-x) = COTX
\frac{ \cos(58) }{ \cos(58) } - \frac{ \sqrt{3 } \cos(38) \sec(38) }{ \ \cot(75) \sqrt{3} \tan(75) }
Using, Cosx × Secx =1
and Cotx × TANX =1
1 - \frac{ \sqrt{3} }{ \sqrt{3} }
= 0

Hope that helps.

Yours Truly
Yumieo


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