Prove? $secx/sinx-cscx/secx=tanx$

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Answer 1 · 2017-11-14T23:49:55

See below:

$secx/sinx-cscx/secx=tanx$

remember that $secx=1/cosx and cscx=1/sinx$

$1/(sinxcosc)-cosx/sinx=tanx$

$1/(sinxcosc)-cosx/sinx(cosx/cosx)=tanx$

$1/(sinxcosc)-cos^2x/(sinxcosx)=tanx$

$(1-cos^2)/(sinxcosc)=tanx$

remember that $sin^2x+cos^2x=1 => sin^2x=1-cos^2x$

$sin^2x/(sinxcosc)=tanx$

$sinx/cosx=tanx$

$tanx=tanx color(white)(000)color(green)root$

Answer 2 · 2017-11-15T09:34:20

$LHS=sec(x)/sin(x) - csc(x)/sec(x)$

$=(sec(x)sec(x) - csc(x)sin(x))/(sinxsecx)$

$=(sec^2(x) - 1)/(sinx/cosx)$

$=tan^2x/tanx$

$= tan(x)=RHS$

Prove? secx/sinx-cscx/secx=tanxsecx/sinx-cscx/secx=tanx