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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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1  
Why the downvote? Perfectly focused question. If this is a homework problem, please tag it with the homework tag. –  Sasha Jul 13 '12 at 21:17
    
You have given us an expression, not an identity. Hint: clear fractions by multiplying numerator and denominator by [ ][ ]. Then use a well known factorisation of the difference of two cubes. –  Mark Bennet Jul 13 '12 at 21:19
    
@Sasha - I didn't down ote, but question would be better posed as how to simplify the expression as there is no identity given. –  Mark Bennet Jul 13 '12 at 21:21
    
crap I am sorry I meant this is equal to 1+sinAcosA –  Fernando Martinez Jul 13 '12 at 21:23
    
@Papigrande I have edited accordingly - try my hint, and note that the expression is equivalent also to $1+\frac 1 2 \sin {2A}$ which I originally thought would be the target expression. - [You should also be concerned about cases where the denominator is zero in the original expression, which will represent removable singularities] –  Mark Bennet Jul 13 '12 at 21:30

3 Answers 3

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.

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With the obvious name changes from where you left off $\frac{c^2/s-s^2/c}{\frac{1}{s}-\frac{1}{c}}= \frac{c^3-s^3}{c-s} = c^2+s^2+cs = 1+cs = 1+ \frac{1}{2}\sin(2A)$

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Hint: $$ \csc(x) = \frac{1}{\sin(x)},\quad \sec(x) = \frac{1}{\cos(x)}$$ So $$\begin{align} \frac{(\cos{A}\cot{A}-\sin{A}\tan{A})}{(\csc{A}-\sec{A})} &= \frac{(\cos{A}\frac{\cos(A)}{\sin(A)}-\sin{A}\frac{\sin(A)}{\cos(A)})\sin(A)\cos(A)}{\cos(A) - \sin(A)} \\ &= \; \dots \end{align} $$

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