# ( 1+ cot(\theta )- csc(\theta ) )( 1- tan(\theta )+ sec(\theta ) )= 2 ?

Feb 17, 2018

The solution to $E q u a t i o n = 2$
is
$\theta = \frac{\pi}{2}$ (modulo $2 \pi$).

HOWEVER, I REALLY THINK THERE IS A MISTAKE IN THE EQUATION AS THINGS WOULD CANCEL OUT NICELY IF THE EQUATION IS THE FOLLOWING:
$E q u a t i o n = \left(1 + \cot \left(\theta\right) - \csc \left(\theta\right)\right) \left(1 + \tan \left(\theta\right) + \sec \left(\theta\right)\right)$

(should be $+ \tan \left(\theta\right)$ instead of $- \tan \left(\theta\right)$).

#### Explanation:

Since the question is vague, I would just try to simplify the equation a bit first.
$E q u a t i o n = \left(1 + \cot \left(\theta\right) - \csc \left(\theta\right)\right) \left(1 - \tan \left(\theta\right) + \sec \left(\theta\right)\right)$

and rewrite everything in terms of sine and cosine:
$= \left(1 + \cos \frac{\theta}{\sin} \left(\theta\right) - \frac{1}{\sin} \left(\theta\right)\right) \left(1 - \sin \frac{\theta}{\cos} \left(\theta\right) + \frac{1}{\cos} \left(\theta\right)\right)$

Simplify a bit more:
$= \left(1 + \frac{\cos \left(\theta\right) - 1}{\sin \left(\theta\right)}\right) \left(1 + \frac{1 - \sin \left(\theta\right)}{\cos \left(\theta\right)}\right)$

Opening up the parentheses:
$= 1 + \frac{1 - \sin \left(\theta\right)}{\cos \left(\theta\right)} + \frac{\cos \left(\theta\right) - 1}{\sin \left(\theta\right)} + \frac{\left(\cos \left(\theta\right) - 1\right) \left(1 - \sin \left(\theta\right)\right)}{\sin \left(\theta\right) \cos \left(\theta\right)}$

And putting everything under the same denominator gives:
$= \frac{\sin \left(\theta\right) \cos \left(\theta\right) + \sin \left(\theta\right) \left(1 - \sin \left(\theta\right)\right) + \cos \left(\theta\right) \left(\cos \left(\theta\right) - 1\right) + \left(\cos \left(\theta\right) - 1\right) \left(1 - \sin \left(\theta\right)\right)}{\sin \left(\theta\right) \cos \left(\theta\right)}$

Now, we look just at the numerator:
$= \sin \left(\theta\right) \cos \left(\theta\right) + \sin \left(\theta\right) - {\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) - \cos \left(\theta\right) + \cos \left(\theta\right) - \sin \left(\theta\right) \cos \left(\theta\right) - 1 + \sin \left(\theta\right)$
$= 2 \sin \left(\theta\right) - {\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) - 1$
$= 2 \sin \left(\theta\right) - {\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) - {\sin}^{2} \left(\theta\right) - {\cos}^{2} \left(\theta\right)$
$= 2 \sin \left(\theta\right) \left(1 - \sin \left(\theta\right)\right)$

Now, back to the equation:
$E q u a t i o n = \frac{2 \sin \left(\theta\right) \left(1 - \sin \left(\theta\right)\right)}{\sin \left(\theta\right) \cos \left(\theta\right)}$
So,
$E q u a t i o n = 2 \frac{\left(1 - \sin \left(\theta\right)\right)}{\cos \left(\theta\right)}$

Now, let's solve for theta when $E q u a t i o n = 2$
This happens when $\frac{1 - \sin \left(\theta\right)}{\cos \left(\theta\right)} = 1$
i.e.
$1 - \sin \left(\theta\right) = \cos \left(\theta\right)$
i.e.
$\sin \left(\theta\right) + \cos \left(\theta\right) = 1$
This only happens when either $\sin \left(\theta\right) = 1$ and $\cos \left(\theta\right) = 0$
when $\theta = \frac{\pi}{2}$ (modulo $2 \pi$)
or when $\sin \left(\theta\right) = 0$ and $\cos \left(\theta\right) = 1$ when $\theta = 0$ (modulo $2 \pi$).
Since $E q u a t i o n = 2 \frac{\left(1 - \sin \left(\theta\right)\right)}{\cos \left(\theta\right)}$, $\cos \left(\theta\right)$ cannot be 0 and
thus the solution to $E q u a t i o n = 2$
is
$\theta = \frac{\pi}{2}$ (modulo $2 \pi$).

HOWEVER, I REALLY THINK THERE IS A MISTAKE IN THE EQUATION AS THINGS WOULD CANCEL OUT NICELY IF THE EQUATION IS THE FOLLOWING:
$E q u a t i o n = \left(1 + \cot \left(\theta\right) - \csc \left(\theta\right)\right) \left(1 + \tan \left(\theta\right) + \sec \left(\theta\right)\right)$
($+ \tan \left(\theta\right)$ instead of $- \tan \left(\theta\right)$).
Then, the equation is indeed 2 for all $\theta$.