Answer:
[tex]\displaystyle{\sin \theta = \dfrac{2ab}{a^2+b^2}}\\\\\displaystyle{\cos \theta = \dfrac{a^2-b^2}{a^2+b^2}}\\\\\displaystyle{\csc \theta = \dfrac{a^2+b^2}{2ab}}\\\\\displaystyle{\sec \theta = \dfrac{a^2+b^2}{a^2-b^2}}\\\\\displaystyle{\cot \theta = \dfrac{a^2-b^2}{2ab}}[/tex]
Step-by-step explanation:
We are given that:
[tex]\displaystyle{\tan \theta = \dfrac{2ab}{a^2-b^2}}[/tex]
To find other trigonometric ratios, first, we have to know that there are total 6 trigonometric ratios:
[tex]\displaystyle{\sin \theta = \sf \dfrac{opposite}{hypotenuse} = \dfrac{y}{r}}\\\\\displaystyle{\cos \theta = \sf \dfrac{adjacent}{hypotenuse} = \dfrac{x}{r}}\\\\\displaystyle{\tan \theta = \sf \dfrac{opposite}{adjacent} = \dfrac{y}{x}}\\\\\displaystyle{\csc \theta = \sf \dfrac{hypotenuse}{opposite} = \dfrac{r}{y}}\\\\\displaystyle{\sec \theta = \sf \dfrac{hypotenuse}{adjacent} = \dfrac{r}{x}}\\\\\displaystyle{\cot \theta = \sf \dfrac{adjacent}{opposite} = \dfrac{x}{y}}[/tex]
Since we are given tangent relation, we know that [tex]\displaystyle{y = 2ab}[/tex] and [tex]\displaystyle{x = a^2-b^2}[/tex], all we have to do is to find hypotenuse or radius (r) which you can find by applying Pythagoras Theorem.
[tex]\displaystyle{r=\sqrt{x^2+y^2}}[/tex]
Therefore:
[tex]\displaystyle{r=\sqrt{(a^2-b^2)^2+(2ab)^2}}\\\\\displaystyle{r=\sqrt{a^4-2a^2b^2+b^4+4a^2b^2}}\\\\\displaystyle{r=\sqrt{a^4+2a^2b^2+b^4}}\\\\\displaystyle{r=\sqrt{(a^2+b^2)^2}}\\\\\displaystyle{r=a^2+b^2}[/tex]
Now we can find other trigonometric ratios by simply substituting the given information below:
Hence:
[tex]\displaystyle{\sin \theta = \dfrac{y}{r} = \dfrac{2ab}{a^2+b^2}}\\\\\displaystyle{\cos \theta = \dfrac{x}{r} = \dfrac{a^2-b^2}{a^2+b^2}}\\\\\displaystyle{\csc \theta = \dfrac{r}{y} = \dfrac{a^2+b^2}{2ab}}\\\\\displaystyle{\sec \theta = \dfrac{r}{x} = \dfrac{a^2+b^2}{a^2-b^2}}\\\\\displaystyle{\cot \theta = \dfrac{x}{y} = \dfrac{a^2-b^2}{2ab}}[/tex]
will be other trigonometric ratios.
