Question

If tan theta = 2ab / a2-b2 then find all other trigonometric ratios ​

Answer 1

`\displaystyle{\sin \theta = (2ab)/(a^2+b^2)}\\\\\displaystyle{\cos \theta = (a^2-b^2)/(a^2+b^2)}\\\\\displaystyle{\csc \theta = (a^2+b^2)/(2ab)}\\\\\displaystyle{\sec \theta = (a^2+b^2)/(a^2-b^2)}\\\\\displaystyle{\cot \theta = (a^2-b^2)/(2ab)}`

Explanation:

We are given that:

`\displaystyle{\tan \theta = (2ab)/(a^2-b^2)}`

To find other trigonometric ratios, first, we have to know that there are total 6 trigonometric ratios:

`\displaystyle{\sin \theta = \sf (opposite)/(hypotenuse) = (y)/(r)}\\\\\displaystyle{\cos \theta = \sf (adjacent)/(hypotenuse) = (x)/(r)}\\\\\displaystyle{\tan \theta = \sf (opposite)/(adjacent) = (y)/(x)}\\\\\displaystyle{\csc \theta = \sf (hypotenuse)/(opposite) = (r)/(y)}\\\\\displaystyle{\sec \theta = \sf (hypotenuse)/(adjacent) = (r)/(x)}\\\\\displaystyle{\cot \theta = \sf (adjacent)/(opposite) = (x)/(y)}`

Since we are given tangent relation, we know that
`\displaystyle{y = 2ab}` and
`\displaystyle{x = a^2-b^2}`, all we have to do is to find hypotenuse or radius (r) which you can find by applying Pythagoras Theorem.

`\displaystyle{r=√(x^2+y^2)}`

Therefore:

`\displaystyle{r=√((a^2-b^2)^2+(2ab)^2)}\\\\\displaystyle{r=√(a^4-2a^2b^2+b^4+4a^2b^2)}\\\\\displaystyle{r=√(a^4+2a^2b^2+b^4)}\\\\\displaystyle{r=√((a^2+b^2)^2)}\\\\\displaystyle{r=a^2+b^2}`

Now we can find other trigonometric ratios by simply substituting the given information below:

• 
  `\displaystyle{x = a^2-b^2}`
• 
  `\displaystyle{y = 2ab}`
• 
  `\displaystyle{r = a^2+b^2}`

Hence:

`\displaystyle{\sin \theta = (y)/(r) = (2ab)/(a^2+b^2)}\\\\\displaystyle{\cos \theta = (x)/(r) = (a^2-b^2)/(a^2+b^2)}\\\\\displaystyle{\csc \theta = (r)/(y) = (a^2+b^2)/(2ab)}\\\\\displaystyle{\sec \theta = (r)/(x) = (a^2+b^2)/(a^2-b^2)}\\\\\displaystyle{\cot \theta = (x)/(y) = (a^2-b^2)/(2ab)}`

will be other trigonometric ratios.