Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.