Verified Answer :

arthurpdc

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 \(\csc x=\dfrac{1}{\sin x}\):

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

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

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

Related Questions