# How to verify this trigonometric identity?

I am having trouble doing this identity.

$$\frac{\cos{A}\cot{A}-\sin{A}\tan{A}}{\csc{A}-\sec{A}} \equiv 1+\cos A\sin A$$

I am stuck I simplified it to.

$$\frac{\cos^{2}{A}\div\sin{A} -\sin^{2}{A}\div\cos{A}}{1\div\sin{A}-1\div\cos{A}}$$

I am in trouble because I know not what to do.

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First, add fractions in the numerator and the denominator, by introducing $1 = \frac{a}{a}$, like so $$\frac{\cos^2(A)}{\sin(A)} - \frac{\sin^2(A)}{\cos(A)} = \frac{\cos^2(A)}{\sin(A)} \cdot \frac{\cos(A)}{\cos(A)} - \frac{\sin^2(A)}{\cos(A)} \cdot \frac{\sin(A)}{\sin(A)} = \frac{\cos^3(A)}{\sin(A) \cos(A)} - \frac{\sin^3(A)}{\sin(A) \cos(A)}$$ Now you can add fractions, since they have the same denominator. Repeat for denominator. And they you should use $(a^3 - b^3) = (a-b)(a^2 + a b + b^2)$ to simplify.
With the obvious name changes from where you left off $\frac{c^2/s-s^2/c}{\frac{1}{s}-\frac{1}{c}}= \frac{c^3-s^3}{c-s} = c^2+s^2+cs = 1+cs = 1+ \frac{1}{2}\sin(2A)$
Hint: $$\csc(x) = \frac{1}{\sin(x)},\quad \sec(x) = \frac{1}{\cos(x)}$$ So \begin{align} \frac{(\cos{A}\cot{A}-\sin{A}\tan{A})}{(\csc{A}-\sec{A})} &= \frac{(\cos{A}\frac{\cos(A)}{\sin(A)}-\sin{A}\frac{\sin(A)}{\cos(A)})\sin(A)\cos(A)}{\cos(A) - \sin(A)} \\ &= \; \dots \end{align}