1+sinx/cosx + cosx/1+sinx = 2secx

Timothy

New member
Joined
Sep 30, 2006
Messages
19
1+sinx/cosx + cosx/1+six = 2secx

1+sin(1+sin)/cos(1+sin) + (cos)cos/cos(1+sin) = 2sec

(1+sin)^2/cos(1+sin) + cos^2/cos(1+sin) =2sec

I am not sure what to do next.

Need help.

Thanks Tim
 
you're getting sloppy ... sin, cos, sec, etc. mean nothing without the x.
also ... use the proper grouping symbols as shown below.

(1+sinx)/cosx + cosx/(1+sinx) = 2secx

common denominator for the left side is cosx(1+sinx) ...

(1+sinx)<sup>2</sup>/[cosx(1+sinx)] + cos<sup>2</sup>x/[cosx(1+sinx)] =

[(1+sinx)<sup>2</sup> + cos<sup>2</sup>x]/[cosx(1+sinx)] =

(1+2sinx+sin<sup>2</sup>x+cos<sup>2</sup>x)/[cosx(1+sinx)] =

(2+2sinx)/[cosx(1+sinx)] =

2(1+sinx)/[cosx(1+sinx)] =

2/cosx = 2secx
 
sinx,cosx

I will work on my sinx and cosx. Thanks for your help.


Tim
 
There is no such a symbol as ‘sinx’
The sine function should written in function notation: sin(x).
The same can be said of the cosine function: cos(x).
Please correct your posting!
Make it readable.
 
Hello, Timothy!

\(\displaystyle \L\frac{1\,+\,\sin x}{\cos x}\,+\,\frac{\cos x}{1\,+\,\sin x}\;=\;2\sec x\)

Multiply top and bottom of the second fraction by \(\displaystyle (1\,-\,sin x):\)

. . \(\displaystyle \L\frac{\cos x}{1\,+\,\sin x}\,\cdot\,\frac{1\,-\,\sin x}{1\,-\,\sin x} \:=\:\frac{\cos x(1\,-\,\sin x)}{1\,-\,\sin^2x} \:=\:\frac{\cos x(1\,-\,\sin x)}{\cos^2x} \:=\:\frac{1\,-\,\sin x}{\cos x}\)


The problem becomes: \(\displaystyle \L\,\frac{1\,+\,\sin x}{\cos x}\,+\,\frac{1\,-\,\sin x}{\cos x}\:=\:\frac{2}{\cos x}\)\(\displaystyle \:=\:2\sec x\)

 
Top