Identity simplification

Question

How do you express $\dfrac{\sin A\sec A\cot A}{\tan A}$ in terms of sine and cosine?

I have simplified using $\sec(A)$ as $\cos^{-1}(A)$ and also $\cot(A)$ as $\dfrac{\cos(A)}{\sin(A)}$, and appear to end up with $\dfrac{\sin(A)}{\cos(A)}$, which is simply $\tan(A)$.

$\sec A$ is not $\cos^{-1}A$; rather, $\sec A$ is $(\cos A)^{-1}$. In standard usage although $\cos^2 A$ means $(\cos A)^2$, nonetheless $\cos^{-1} A$ does not mean $(\cos A)^{-1}$. Rather it means the inverse function of the cosine function, evaluated at $A$. ${}\qquad{}$ — Michael Hardy, Jun 11 '15 at 14:30

score 2 · Answer 1 · edited Jun 11 '15 at 14:29

2

$$\sin(A)\frac{\sec(A)}{\tan{A}}\cot{A}=\frac{\sin(A)}{\cos(A)}\frac{1}{\tan(A)}\cot(A)=\cot(A)=\frac{\cos(A)}{\sin(A)}$$

edited Jun 11 '15 at 14:29

Michael Hardy

1

answered Jun 11 '15 at 14:18

Joaquin Liniado

3,840

score 0 · Answer 2 · answered Jun 11 '15 at 14:43

$$ \frac{\sin A\sec A\cot A}{\tan A} = \frac{\sin A\cdot\dfrac 1 {\cos A}\cdot \dfrac {\cos A} {\sin A}}{\dfrac{\sin A}{\cos A}} $$

That expresses it in terms of sine and cosine. After that, you can do some simplifying. Dividing by $\sin A/\cos A$ is the same as multiplying by $\cos A/\sin A$, so you get $$ \sin A\cdot\dfrac 1 {\cos A}\cdot \dfrac {\cos A} {\sin A} \cdot \frac{\cos A}{\sin A}. $$ Then, doing a couple of cancelations of factors common to the numerator and the denominator, you get $$ \frac{\cos A}{\sin A}. $$ That expresses it in terms of the sine and the cosine of $A$.

You can also write it as $\cot A$.

No...somehow the initial question got put in wrong. Sin A Sec A are in the numerator....and tan A Cot A are in the denominator. — greg, Jun 11 '15 at 16:10

Identity simplification

2 Answers2