How do you prove $(sinx+cosx)/sinx - (cosx-sinx)/cosx = sec x csc x$ ?

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Answer 1 · 2018-04-24T16:14:13

Please see below.

We know that,

$color(red)((1)cos^2x+sin^2x=1$

$color(blue)((2)1/cosx=secx and 1/sinx=cscx$

Here,

$(sinx+cosx)/sinx-(cosx-sinx)/cosx=secxcscx$

We take,

$LHS=(sinx+cosx)/sinx-(cosx-sinx)/cosx$

$=(cosx(sinx+cosx)-sinx(cosx-sinx))/(sinxcosx)$

$=(cancel(sinxcosx)+cos^2x-cancel(sinxcosx)+sin^2x)/(sinxcosx$

$=color(red)((cos^2x+sin^2x))/(sinxcosx)...tocolor(red)(Apply(1)$

$=color(red)(1)/(sinxcosx)$

$=color(blue)(1/cosx*1/sinx...tocolor(blue)(Apply(2)$

$=color(blue)(secxcscx$

$=RHS$

How do you prove (sinx+cosx)/sinx - (cosx-sinx)/cosx = sec x csc x(sinx+cosx)/sinx - (cosx-sinx)/cosx = sec x csc x?