More Proofs

[lola]

(tan x/1+sec x) + (1+sec x/tan x) = 2csc x (this is the proof im working on now)


I'm not sure what to do on this one either......
Here is what i got

I got a common denominator first:

[(tan^2 x +1)+(sec^2 x)]/tan x(1+sec x)

so i remember about the identity:
tan^2 x + 1 = sec^2 x ---> so I substitute

(sec^2 x+ sec^2 x)/ tan x(1+sec x)

now i replaced everything with sines and cosines...
[(1/cos^2 x) + (1/cos^2 x)] / [(sin x/cos x)(1+(1/cos x))]

is it now time to cross cancel or is this one not possible

 

[lola]

Need help with a proof

(tan x/1+sec x) + (1+sec x/tan x) = 2csc x (this is the proof im working on now)


I'm not sure what to do on this one either......
Here is what i got

I got a common denominator first:

[(tan^2 x +1)+(sec^2 x)]/tan x(1+sec x)

so i remember about the identity:
tan^2 x + 1 = sec^2 x ---> so I substitute

(sec^2 x+ sec^2 x)/ tan x(1+sec x)

now i replaced everything with sines and cosines...
[(1/cos^2 x) + (1/cos^2 x)] / [(sin x/cos x)(1+(1/cos x))]

is it now time to cross cancel or is this one not possible

 

[ChaoticLlama]

\(\displaystyle \begin{array}{l}
\frac{{\tan x}}{{1 + \sec x}} + \frac{{1 + \sec x}}{{\tan x}} \\
= \frac{{\tan ^2 x + (1 + \sec x)^2 }}{{(1 + \sec x)\tan x}} \\
= \frac{{\tan ^2 x + 1 + 2\sec x + \sec ^2 x}}{{(1 + \sec x)\tan x}} \\
= \frac{{\sec ^2 x - 1 + 1 + 2\sec x + \sec ^2 x}}{{(1 + \sec x)\tan x}} \\
= \frac{{2\sec ^2 x + 2\sec x}}{{(1 + \sec x)\tan x}} \\
\end{array}\)

can you go from here?