# Math Help - Trigonometry functions

1. ## Trigonometry functions

Could someone please explain these 2 problems for me?

#1: If $0 and $sinx + cosx = \frac{1}{5}$, then find tan x.

#2: What are the solutions for $sec[arctan(\frac{-3}{5})]$ and $sin[arcsin(\frac{5}{13}) - arccos(\frac{-3}{5})]$?

Thank you!!!

2. #1: If $0 and $sinx + cosx = \frac{1}{5}$, then find tan x.

Not sure about this, but I reckon you have to change sinx + cosx into the form Rsin(x + theta)

$Rsin(x+\theta)=Rsinxcos\theta+Rcosxsin\theta" alt="Rsin(x+\theta)=Rsinxcos\theta+Rcosxsin\theta" />
so $Rsinxcos\theta+Rcosxsin\theta=sinx+cosx$
Equating terms $Rsinxcos\theta=sinx$
and $Rcosxsin\theta=cosx$

hence $Rcos\theta=1$ and $Rsin\theta=1$
so $tan\theta=1,\theta=\frac{\pi}{4}$

then find $R=\frac{1}{\sqrt{2}}$

so you will have $\frac{1}{\sqrt{2}}sin(x+\frac{\pi}{4})=\frac{1}{5}$

hope this helps, it should be easy to find x now and therefore find tan x.... there is probably an easier way though!!