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write in terms of sin and cos: (sin(x)sec(x))/(tan(x))
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\[\frac{ \sin(x) }{ \sin(x)/\cos(x) }\times \frac{ 1/\cos(x) }{ \sin(x)/\cos(x) }\]
\[(\sin (x) \times \frac{ \sin(x) }{ \cos(x) }) \times (\frac{ 1/co(x) }{ \sin(x)/\cos(x) })\]
\[\frac{ \sin ^{2}(x) }{ \cos(x) } \times \frac{ \sin(x) }{ \cos(x) }\]
\[\frac{ \sin ^{3}x }{ \cos x }\]
i think that is the final answer
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I think it is this: \[\frac{ sinx*\frac{ 1 }{ cosx } }{ \frac{ sinx }{ cosx } }\] Now put everything together: \[\frac{ \frac{ sinx }{ cosx } }{ \frac{ sinx }{ cosx } } = 1\] ZeHanz
You could also say: sin(x)sec(x) = sin(x)/cos(x) = tan(x). Put this in the expression, which then reads: tan(x) / tan(x) = 1. Hope this helps. ZeHanz
thanks for your time
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