Question

Math Analysis: Simplify each expression(cos x/1+sinx) + (1+sinx/cosx)

Answer 1

\[1+\csc x = 1+\frac{1}{\sin x} = \frac{1+ \sin x}{\sin x}\]
\[\frac{\cos x}{1+csc x} = \cos{x}*\frac{\sin x}{1+\sin x} = \frac{\sin x \cos x}{1+\sin
x}\]
\[\frac{\sin x \cos x}{1+\sin x} +\frac{1+\sin x}{\cos x} = \frac{\sin x \cos^{2} x + (1+\sin x)^{2}}{\cos x(1+\sin x)} = \frac{\sin x (1-\sin^{2} x) + (1+\sin x)^{2}}{\cos x(1+\sin x)}\]
\[=\frac{\sin x (1-\sin x)(1+\sin x) + (1+\sin x)^{2}}{\cos x(1+\sin x)}\]
\[=\frac{(1+\sin x)(\sin x(1-\sin x) + (1+\sin x))}{\cos x(1+\sin x)}\]
\[=\frac{(\sin x(1-\sin x) + (1+\sin x))}{\cos x}\]
\[=\frac{1+2\sin x -\sin^{2} x}{\cos x}\]
\[=\frac{2\sin x + \cos^{2} x}{\cos x} = 2\tan x +\cos x\]

Answer 2

oops sorry its (cos x/1+sinx) + (1+sin x/cosx) lol o-o

Answer 3

wow big screw up there..haha
um well pretty much the same steps except for the beginning

Answer 4

uh, soo how would i start this off?

Answer 5

\[\frac{\cos x}{1+\sin x}+\frac{1+\sin x}{\cos x} = \frac{\cos^{2} +(1+\sin x)^{2}}{\cos x(1+\sin x)}\]

Answer 6

ok then what

Answer 7

i got that part so far

Answer 8

expand numerator and simplify

Answer 9

\[(1+\sin x)^{2} = \sin^{2} x +2\sin x +1\]

Answer 10

ok so i get. cos^2x + 1 + sinx + sinx + sin^2x/cosx(1+sinx)

Answer 11

trig identity
sin^2 +cos^2 = 1

Answer 12

so i now get 2+2sinx/cosx(1+sinx)

Answer 13

yep, factor out the 2 ...

Answer 14

OHHH

Answer 15

OKAY TY XD

Answer 16

:]]

Answer 17

:)

Answer 18

that was hard but easy at the same time lol