Question

If $\sin \left(\mathrm{\theta }\right)=\frac{2}{3}$ find all other trigonometric ratios?

Answer 1

Step 1: Solve for the value of $\cos \left(\mathrm{\theta }\right)$

Given, $\sin \left(\mathrm{\theta }\right)=\frac{2}{3}$

$\begin{array}{ccc}& \cos \left(\mathrm{\theta }\right)=\sqrt{1-{\sin }^2\left(\mathrm{\theta }\right)}& ;\left[\because {\sin }^2\left(\mathrm{\theta }\right)+{\cos }^2\left(\mathrm{\theta }\right)=1\right]\\ \Rightarrow & \cos \left(\mathrm{\theta }\right)=\sqrt{1-\frac{4}{9}}& \\ \Rightarrow & \cos \left(\mathrm{\theta }\right)=\frac{\sqrt{5}}{3}& \end{array}$

Step 2: Solve for the value of $\tan \left(\mathrm{\theta }\right)$

$\begin{array}{ccc}& \tan \left(\mathrm{\theta }\right)=\frac{\sin \left(\mathrm{\theta }\right)}{\cos \left(\mathrm{\theta }\right)}& \\ \Rightarrow & \tan \left(\mathrm{\theta }\right)=\frac{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{{\displaystyle \raisebox{1ex}{$\sqrt{5}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}& \\ \Rightarrow & \tan \left(\mathrm{\theta }\right)=\frac{2}{\sqrt{5}}& \end{array}$

Step 3: Solve for the value of $\sec \left(\theta \right)$

$\begin{array}{ccc}& \sec \left(\mathrm{\theta }\right)=\frac{1}{\cos \left(\mathrm{\theta }\right)}& \\ \Rightarrow & \sec \left(\mathrm{\theta }\right)=\frac{3}{\sqrt{5}}& \end{array}$

Step 4: Solve for the value of $\csc \left(\theta \right)$

$\begin{array}{ccc}\Rightarrow & cosec\left(\mathrm{\theta }\right)=\frac{1}{\sin \left(\mathrm{\theta }\right)}& \\ \Rightarrow & cosec\left(\mathrm{\theta }\right)=\frac{3}{2}& \end{array}$

Step 5: Solve for the value of $\cot \left(\theta \right)$

$\begin{array}{ccc}& \cot \left(\mathrm{\theta }\right)=\frac{1}{\tan \left(\mathrm{\theta }\right)}& \\ \Rightarrow & \cot \left(\mathrm{\theta }\right)=\frac{\sqrt{5}}{2}& \end{array}$

Hence, the trigonometric ratios when $\sin \left(\mathrm{\theta }\right)=\frac{2}{3}$ are:

$\begin{array}{rcl}\cos \left(\mathrm{\theta }\right)& =& \frac{\sqrt{5}}{3}\\ \tan \left(\mathrm{\theta }\right)& =& \frac{2}{\sqrt{5}}\\ \csc \left(\mathrm{\theta }\right)& =& \frac{3}{2}\\ \sec \left(\mathrm{\theta }\right)& =& \frac{3}{\sqrt{5}}\\ \cot \left(\mathrm{\theta }\right)& =& \frac{\sqrt{5}}{2}\end{array}$