Question #d8d7c

Question

1570 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-04T16:58:09

$(1 + tan x + sec x) (1 + cot x - csc x)$ $(1+tanx+secx)(1+cotx-cscx)$

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

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

$= \frac{{(cos x + sin x)}^{2} - 1}{cos x sin x}$ $=((cosx+sinx)^2-1)/(cosxsinx)$

$= \frac{{cos}^{2} x + {sin}^{2} x + 2 sin x cos x - 1}{cos x sin x}$ $=(cos^2x+sin^2x+2sinxcosx-1)/(cosxsinx)$

$= \frac{1 + 2 sin x cos x - 1}{cos x sin x}$ $=(1+2sinxcosx-1)/(cosxsinx)$

$= \frac{2 sin x cos x}{cos x sin x}$ $=(2sinxcosx)/(cosxsinx)$

$= 2$ $=2$