How do you simplify tan^2x(csc^2x-1)?

Jul 26, 2015

By using the Trigonometric Identity : ${\sin}^{2} x + {\cos}^{2} x = 1$

Explanation:

Divide both sides of the above identity by ${\sin}^{2} x$ to obtain,

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

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

$\implies 1 + \frac{1}{\tan} ^ 2 x = {\csc}^{2} x$

$\implies {\csc}^{2} x - 1 = \frac{1}{\tan} ^ 2 x$

Now, we are able to write : ${\tan}^{2} x \left({\csc}^{2} x - 1\right) \text{ }$ as $\text{ } {\tan}^{2} x \left(\frac{1}{\tan} ^ 2 x\right)$

and the result is $\textcolor{b l u e}{1}$

Jul 27, 2015

Simplify: ${\tan}^{2} x \left({\csc}^{2} x - 1\right)$

Explanation:

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

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