How do you solve $(1 + sinx + cosx)/(1 + sinx - cosx) = (1 + cosx)/sinx$ ?

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Answer 1 · 2018-04-28T14:30:31

$LHS=(1+sinx+cosx)/(1+sinx-cosx)$

$=(sinx/sinx)*(1+sinx+cosx)/(1+sinx-cosx)$

$=1/sinx[(sinx+sin^2x+sinx*cosx)/(1+sinx-cosx)]$

$=1/sinx[(sinx(1+cosx)+(1+cosx)(1-cosx))/(1+sinx-cosx)]$

$=1/sinx[((1+cosx)cancel((sinx+1-cosx)))/(cancel((sinx+1-cosx)]$

$=(1+cosx)/sinx=RHS$

How do you solve (1 + sinx + cosx)/(1 + sinx - cosx) = (1 + cosx)/sinx(1 + sinx + cosx)/(1 + sinx - cosx) = (1 + cosx)/sinx?