How do you verify (tan^2x + 1) / tan^2x = csc^2x?

Jun 19, 2016

Use the Pythagorean identity ${\tan}^{2} x + 1 = {\sec}^{2} x$ to start the simplification on the left side.

Explanation:

${\sec}^{2} \frac{x}{\tan} ^ 2 x =$

Recall that ${\sec}^{2} x = \frac{1}{\cos} ^ 2 x$, and that ${\tan}^{2} x = {\sin}^{2} \frac{x}{\cos} ^ 2 x$

$\frac{\frac{1}{\cos} ^ 2 x}{{\sin}^{2} \frac{x}{{\cos}^{2} x}} =$

$\frac{1}{\cos} ^ 2 x \times {\cos}^{2} \frac{x}{\sin} ^ 2 x =$

$\frac{1}{\sin} ^ 2 x =$

Remember that $\frac{1}{\sin} x = \csc x$

${\csc}^{2} x =$

Identity proved!!!

Hopefully this helps!

Jun 19, 2016

The definition of tangent is such that

$\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$, then the left side of the equation is

$\frac{\left(\sin {\left(x\right)}^{2} / \cos {\left(x\right)}^{2}\right) + 1}{\sin {\left(x\right)}^{2} / \cos {\left(x\right)}^{2}}$

$= \left(\left(\sin {\left(x\right)}^{2} / \cos {\left(x\right)}^{2}\right) + 1\right) \cdot \cos {\left(x\right)}^{2} / \sin {\left(x\right)}^{2}$

$= 1 + \cos {\left(x\right)}^{2} / \sin {\left(x\right)}^{2}$

$= \frac{\sin {\left(x\right)}^{2} + \cos {\left(x\right)}^{2}}{\sin} {\left(x\right)}^{2}$

and, using the fundamental property of $\sin$ and $\cos$ that $\sin {\left(x\right)}^{2} + \cos {\left(x\right)}^{2} = 1$ we have

$\frac{\sin {\left(x\right)}^{2} + \cos {\left(x\right)}^{2}}{\sin {\left(x\right)}^{2}} = \frac{1}{\sin {\left(x\right)}^{2}} = \csc {\left(x\right)}^{2}$.