How do you solve $\frac{1 + sin x + cos x}{1 + sin x - cos x} = \frac{1 + cos x}{sin x}$ $(1 + sinx + cosx)/(1 + sinx - cosx) = (1 + cosx)/sinx$ ?

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Answer 1 · 2018-04-28T14:30:31

$L H S = \frac{1 + sin x + cos x}{1 + sin x - cos x}$ $LHS=(1+sinx+cosx)/(1+sinx-cosx)$

$= (\frac{sin x}{sin x}) \cdot \frac{1 + sin x + cos x}{1 + sin x - cos x}$ $=(sinx/sinx)*(1+sinx+cosx)/(1+sinx-cosx)$

$= \frac{1}{sin x} [\frac{sin x + {sin}^{2} x + sin x \cdot cos x}{1 + sin x - cos x}]$ $=1/sinx[(sinx+sin^2x+sinx*cosx)/(1+sinx-cosx)]$

$= \frac{1}{sin x} [\frac{sin x (1 + cos x) + (1 + cos x) (1 - cos x)}{1 + sin x - cos x}]$ $=1/sinx[(sinx(1+cosx)+(1+cosx)(1-cosx))/(1+sinx-cosx)]$

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

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

How do you solve (1 + sinx + cosx)/(1 + sinx - cosx) = (1 + cosx)/sinx1+sinx+cosx1+sinx−cosx=1+cosxsinx(1 + sinx + cosx)/(1 + sinx - cosx) = (1 + cosx)/sinx?