Trigonometric functions of the acute angle Find the other five trigonometric functions of the acute angle A, given that:
\begin{align}
&\text{a)}\ \ \sec A = 2  \\[15pt]
&\text{b)}\ \ \cos A = \frac{m^2 - n^2}{m^2 + n^2}
\end{align}
Help me. I don't know how to solve this one. Thanks.
 A: It is useful to know at least some basic trigonometric identities. See List of trigonometric identities at Wikipedia for a very complete list.

a) Since $\cos A=\frac1{\sec A}$, you get $\cos A=\frac12$. Can you get possible values of $A$ from there?

b) If $$\cos A=\frac{m^2-n^2}{m^2+n^2}=\left(\frac m{\sqrt{m^2+n^2}}\right)^2-\left(\frac n{\sqrt{m^2+n^2}}\right)^2,$$ then for an angle $B$ such that $\cos B=\frac m{\sqrt{m^2+n^2}}$ and $\sin B=\frac m{\sqrt{m^2+n^2}}$ you get $$\cos A=\cos^2B-\sin^2B=\cos2B.$$ Maybe this could help to express $A$ using $B$. You should be able to find $B$ using some inverse trigonometric functions. Double angle formulae should help you to get the values of $\sin A$, $\tan A$, etc for $A=2B$. (You should also think about the possibility $A=-2B$. Do you know why?)
A: for case a)
$$\sec A = 2$$
$$\cos A=\frac{1}{\sec A}=\frac{1}{2}$$
$$\sin A=\sqrt{1-\cos^2A}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$$
$$\tan A=\frac{\sin A}{\cos A}=\sqrt 3$$
$$\operatorname{cotg} A=\frac{1}{\tan A}=\frac{1}{\sqrt 3}$$
and for b) follow Martin Sleziak
A: Honestly, I suspect that equation b is a red herring.
\begin{align}
\cos{A}
&=\frac{1}{\sec{A}}\\
&=\frac{1}{2}
\end{align}
\begin{align}
\sin{A}
&=\sqrt{1-\cos^2{A}}\\
&=\frac{\sqrt{3}}{2}
\end{align}
\begin{align}
\tan{A}
&=\frac{\sin{A}}{\cos{A}}\\
&=\sqrt{3}
\end{align}
\begin{align}
\csc{A}
&=\frac{1}{\sin{A}}\\
&=\frac{2}{\sqrt{3}}\\
\end{align}
\begin{align}
\cot{A}
&=\frac{1}{\tan{A}}\\
&=\frac{1}{\sqrt{3}}\\
\end{align}
A: $$\sec A = 2$$
using an uncommon Pythagorean Identity$\dots$
$$\sec^2A+\csc^2A=\sec^2A\csc^2A$$
$$\csc^2A = \frac{\sec^2A}{\sec^2A-1}=4/3\therefore\csc A=2/\sqrt{3}$$
This allows us to build this right triangle.

from the triangle with opposite, adjacent, and hypotenuse sides $2, 2/\sqrt{3}, 4/\sqrt{3}$ we have $\dots$
$$\sin A = \frac{\sec A}{\sec A\csc A}=\frac{1}{\csc A}=\sqrt{3}/2$$
similarly
$$\cos A = \frac{\csc A}{\sec A\csc A}=\frac{1}{\sec A}=1/2$$
lastly
$$\tan A = \frac{\sec A}{\csc A}=\frac{2}{2/\sqrt{3}}=\sqrt{3}$$
and
$$\cot A = \frac{\csc A}{\sec A}=\frac{2/\sqrt{3}}{2}=1/\sqrt{3}$$
A: I would assume that this is actually 2 different problems.
Try doing this:
Draw a right triangle, and assign values to 2 of the sides. In the case of $\sec A=2=\frac{2}{1}$, draw a triangle where the hypotenuse is 2 and the adjacent side is 1. Then use the Pythagorean theorem to find the other side (in this case it is $\sqrt{2^2-1^2}=\sqrt{3}$). You can then just read off the trigonometric values from the triangle.

