# Thread: 2 questions: proving identity and reducing

1. ## 2 questions: proving identity and reducing

The first question I think I have the correct answer to but wanted to double check.

The problem says prove the following identity...

sec x + csc x = csc x + csc x tan x

(1/cosx + 1/sin)=

sinx/(sinxcosx) + cosx/(sinxcosx)

(1/sinx)*(sinxcosx)= cscxtanx=secx

The second question I have I need to reduce this to a single term.

tan (x+y) - tan y
-----------------------
1+ tan (x+y) tan y

I know it reduces to tanx, but how would I show the work for a problem like this?

Thanks so much for any help!

2. Originally Posted by KarlosK
The first question I think I have the correct answer to but wanted to double check.

The problem says prove the following identity...

sec x + csc x = csc x + csc x tan x

(1/cosx + 1/sin)=

sinx/(sinxcosx) + cosx/(sinxcosx)

(1/sinx)*(sinxcosx)= cscxtanx=secx

Does this answer make sense? no

working from the right side ...

$\textcolor{red}{\csc{x} + \csc{x}\tan{x}}$

$\textcolor{red}{\csc{x} + \frac{1}{\sin{x}} \cdot \frac{\sin{x}}{\cos{x}}}$

$\textcolor{red}{\csc{x} + \frac{1}{\cos{x}}}$

$\textcolor{red}{\csc{x} + \sec{x}}$

The second question I have I need to reduce this to a single term.

tan (x+y) - tan y
-----------------------
1+ tan (x+y) tan y

use this identity ...

$\textcolor{red}{\tan(a-b) = \frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}}$

let $\textcolor{red}{a = (x+y)}$ and $\textcolor{red}{b = y}$
...

3. Skeeter thanks for the help on the first problem, I now understand how you did that. On the second problem I am still a little unclear as to how I should show the work to the professor.

So Tan A-B = x+y--y = x

tan x.

Is that enough?

4. $\frac{\tan(x+y) - \tan{y}}{1 + \tan(x+y)\tan{y}} = \tan[(x+y) - y] = \tan{x}$