1. ## trigo indentity

pls help me with this question, been working on it for quite some time but still can't figure it out:

Given cosecA+cotA=3, evaluate cosecA-cotA and cosA.

2. $\frac{1}{\sin A}+\frac{\cos A}{\sin A} = 3$
Thus,
$1+\cos A = 3\sin A$
Square,
$1+2\cos A + \cos^2 A = 3\sin^2 A$
Thus,
$1+2\cos A+\cos^2 A = 3(1-\cos^2 A)$
Solve for,
$\cos A$.

3. $
\csc(A)+\cot(A) = \frac{1}{\sin(A)}+\frac{\cos(A)}{\sin(A)}
$

$
= \frac{1+\cos(A)}{\sin(A)}
$

$
= \frac{(1+\cos(A))(1-\cos(A))}{\sin(A)(1-\cos(A))}
$

$
= \frac{1-\cos^{2}(A)}{\sin(A)(1-\cos(A))}
$

$
= \frac{\sin^{2}(A)}{\sin(A)(1-\cos(A))}
$

$
= \frac{\sin(A)}{1-\cos(A)}
$

$
= \frac{1}{\frac{1-\cos(A)}{\sin(A)}}
$

$
= \frac{1}{\frac{1}{\sin(A)}-\frac{\cos(A)}{\sin(A)}}
$

$
= \frac{1}{\csc(A)-\cot(A)}
$

I suspect it's cleaner if you remember relationships between the cosecant and cotangent, but I never do. I always have to look them up.

Okay, you do cos(A).

Given: . $\csc A + \cot A \:=\:3$, .evaluate: . $\csc A - \cot A$

Note that: . ${\color{blue}A \neq 0,\,\pi,\,\cdots}$
We have: . $\csc A + \cot A \:=\:3$

Multiply by $\frac{\csc A - \cot A}{\csc A - \cot A}\!:\;\;\frac{\csc A + \cot A}{1}\cdot\frac{\csc A - \cot A}{\csc A - \cot A} \;=\;3$

. . $\frac{\csc^2\!A - \cot^2\!A}{\csc A - \cot A} \:=\:3\quad\Rightarrow\quad\frac{1}{\csc A - \cot A} \:=\:3$

Therefore: . $\csc A - \cot A \:=\:\frac{1}{3}$

Evaluate: $\cos A$

We have: . $\frac{1}{\sin A} + \frac{\cos A}{\sin A} \:=\:3\quad\Rightarrow\quad1 + \cos A \:=\:3\sin A$

Square: . $(1 + \cos A)^2 \:=\:9\sin^2\!A\quad\Rightarrow\quad1 + 2\cos A + \cos^2\!A \:=\:9(1-\cos^2\!A)$

. . which simplifies to: . $5\cos^2A + \cos A - 4 \:=\:0\quad\Rightarrow\quad(\cos + 1)(5\cos A - 4) \:=\:0$

and we have: . $\begin{array}{cccccccccc}\cos A + 1 & = & 0 & \Rightarrow & \cos A & = & \text{-}1 & \Rightarrow & A \: = \:\pi & \text{extraneous} \\
6\cos A - 4 & = & 0 & \Rightarrow & \cos A & = & \frac{4}{5} & \Rightarrow & A \: = \:\cos^{\text{-}1}\!\left(\frac{4}{5}\right) &
\end{array}$

5. thx for the different ways of going about solving the qns, i understand it much better now