# How do you verify csc^2xcot^2x+csc^2x=csc^4x ?

Mar 5, 2018

Take ${\csc}^{2} \left(x\right)$ common then it becomes:

${\csc}^{2} \left(x\right)$(1+ ${\cot}^{2} \left(x\right)$)

which is equal to ${\csc}^{4} \left(x\right)$

#### Explanation:

Basic trigonometric identities To prove $\left({\csc}^{2} x {\cot}^{2} x + {\csc}^{2} x\right) = {\csc}^{4} x$

Taking common term ${\csc}^{2} x$ on L H S out,

$\implies {\csc}^{2} x \left(1 + {\cot}^{2} x\right)$

But ${\csc}^{2} x = 1 + {\cot}^{2} x$ (Trigonometric identity.)

Hence $\implies {\csc}^{2} x \cdot {\csc}^{2} x = {\csc}^{4} x = R H S$

Q E D.

Mar 5, 2018

It is given that, ${\csc}^{2} x {\cot}^{2} x + {\csc}^{2} x = {\csc}^{4} x$

$L H S \implies \frac{1}{\sin} ^ 2 x \cdot {\cos}^{2} \frac{x}{\sin} ^ 2 x + \frac{1}{\sin} ^ 2 x$

${\cos}^{2} \frac{x}{\sin} ^ 4 x + {\sin}^{2} \frac{x}{\sin} ^ 4 x$

$\frac{{\cos}^{2} x + {\sin}^{2} x}{\sin} ^ 4 x$

$\frac{1}{\sin} ^ 4 x = {\csc}^{4} x$; [as we know color(red)(sin^2x + cos^2x = 1]