Question

Verify that the following equation is an identity. (1/sinx) - (1/cscx) = cscx - sinx

Answer 1

$\mathrm{csc}\,x = \frac{1}{\sin x}$, so:

$\dfrac{1}{\sin x}-\dfrac{1}{\mathrm{csc}\,x} = \dfrac{1}{\sin x} - \frac{1}{\frac{1}{\sin x}} = \dfrac{1}{\sin x} - \sin x = \mathrm{csc}\,x -\sin x \quad \square$

Answer 2

Manipulating the left side of the equation, we obtain:

$\dfrac{1}{\sin x}-\dfrac{1}{\csc x}=\dfrac{\csc x-\sin x}{\sin x\cdot\csc x}$

Using that $\csc x=\dfrac{1}{\sin x}$:

$\dfrac{\csc x-\sin x}{\sin x\cdot\csc x}=\dfrac{\csc x-\sin x}{\sin x\cdot\dfrac{1}{\sin x}}=\dfrac{\csc x-\sin x}{1}=\csc x-\sin x\\\\\boxed{\dfrac{1}{\sin x}-\dfrac{1}{\csc x}=\csc x-\sin x}~~\blacksquare$