+6250 \(\cot(\theta)=\dfrac 1 4 \\ \dfrac {\cos(\theta)}{\sin(\theta)}=\dfrac 1 4 \\ \dfrac{\sqrt{1-\sin^2(\theta)}}{\sin(\theta)}=\dfrac 1 4\\ \dfrac{1-\sin^2(\theta)}{\sin^2(\theta)}=\dfrac 1 {16} \\ 16 -16\sin^2(\theta)=\sin^2(\theta)\)
\(16 = 17\sin^2(\theta) \\ \sin^2(\theta)=\dfrac {16}{17} \\ \sin(\theta)=\sqrt{\dfrac{16}{17}}=\dfrac{4}{\sqrt{17}}\)
\(\cos(\theta)=\sqrt{1-\sin^2(\theta)} = \sqrt{1-\left(\dfrac{4}{\sqrt{17}}\right)^2}=\sqrt{\dfrac {1}{17}}\\ \sec(\theta)=\dfrac 1 {\cos(\theta)} = \sqrt{17}\)
.+6250 \(\cot(\theta)=\dfrac 1 4 \\ \dfrac {\cos(\theta)}{\sin(\theta)}=\dfrac 1 4 \\ \dfrac{\sqrt{1-\sin^2(\theta)}}{\sin(\theta)}=\dfrac 1 4\\ \dfrac{1-\sin^2(\theta)}{\sin^2(\theta)}=\dfrac 1 {16} \\ 16 -16\sin^2(\theta)=\sin^2(\theta)\)
\(16 = 17\sin^2(\theta) \\ \sin^2(\theta)=\dfrac {16}{17} \\ \sin(\theta)=\sqrt{\dfrac{16}{17}}=\dfrac{4}{\sqrt{17}}\)
\(\cos(\theta)=\sqrt{1-\sin^2(\theta)} = \sqrt{1-\left(\dfrac{4}{\sqrt{17}}\right)^2}=\sqrt{\dfrac {1}{17}}\\ \sec(\theta)=\dfrac 1 {\cos(\theta)} = \sqrt{17}\)
+118629 Thanks Rom,
When I do these i start by drawing a right anged triangle and mark one of the acute angels as theta.
sine cot theta = 1/4
opposite will be 4 and adjacent will be 1
use pythagoras to get the third side sqrt(1+16) = sqrt{(17)
Now you can read all th other ratios off the triangle.
This only gives you the ratios for the acute angles though.
Since cot is positive, tan is positive so theta can be in the 1st OR third quadrants.
So some of the ratios can be + or -
:))
