Hello,
I am having a bit of trouble solving this eqn:
tan (x) + cot (x) = 2sec (x)
For all angles between 0 and 360 degrees.
Thank you!
Max
$\displaystyle tanx+cotx=\frac{sinx}{cosx}+\frac{cosx}{sinx}=\fra c{sinx}{sinx}\ \frac{sinx}{cosx}+\frac{cosx}{cosx}\ \frac{cosx}{sinx}$
$\displaystyle =\frac{sin^2x+cos^2x}{sinxcosx}$
$\displaystyle \Rightarrow\ \frac{1}{sinxcosx}=\frac{2}{cosx}\ \Rightarrow\ \frac{1}{sinx}=2\ \Rightarrow\ sinx=\frac{1}{2}$
The vertical co-ordinate of the angle is 0.5 in the unit circle, centre (0,0).
This corresponds to two angles,
$\displaystyle x=sin^{-1}\left(\frac{1}{2}\right),\ x={\pi}-sin^{-1}\left(\frac{1}{2}\right)$
$\displaystyle x=30^o,\ x=150^o$
$\displaystyle \frac{sinx}{cosx}+\frac{3cosx}{sinx}=\frac{5}{cosx }$
$\displaystyle \frac{sin^2x+3cos^2x}{cosxsinx}=\frac{5}{cosx}$
$\displaystyle \frac{sin^2x+cos^2x+2cos^2x}{cosxsinx}=\frac{1+2co s^2x}{cosxsinx}=\frac{5}{cosx}$
$\displaystyle \frac{1+2cos^2x}{sinx}=5$
$\displaystyle 2cos^2x+1=5sinx$
$\displaystyle 2-2sin^2x-5sinx+1=0$
$\displaystyle 2sin^2x+5sinx-1=0$
$\displaystyle sinx=\frac{-5\pm\sqrt{25+8}}{4}$
x is the inverse sine of the 2 results.