1. ## Trigonometry functions

Could someone please explain these 2 problems for me?

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

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

Thank you!!!

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

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

$\displaystyle Rsin(x+\theta)=Rsinxcos\theta+Rcosxsin\theta$
so $\displaystyle Rsinxcos\theta+Rcosxsin\theta=sinx+cosx$
Equating terms $\displaystyle Rsinxcos\theta=sinx$
and $\displaystyle Rcosxsin\theta=cosx$

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

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

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

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