# Simplify the expression (sin(a)cos(b)+cos(a)sin(b))/(cos(a)cos(b)-sin(a)sin(b)) * (cos(a)cos(b)+sin(a)sin(b))/(sin(a)cos(b)-cos(a)sin(b))?

Apr 27, 2017

$\tan \left(a + b\right) \cdot \cot \left(a - b\right)$

#### Explanation:

The expression is:

 E=(sin(a)cos(b)+cos(a)sin(b))/(cos(a)cos(b)-sin(a)sin(b)) * (cos(a)cos(b)+sin(a)sin(b))/(sin(a)cos(b)-cos(a)sin(b))

We can us the sine and cosine sum identities:

$\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B$
$\sin \left(A - B\right) = \sin A \cos B - \cos A \sin B$
$\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$
$\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$

Applying these identities we can rewrite the expression as:

$E = \frac{\sin \left(a + b\right)}{\cos \left(a + b\right)} \cdot \frac{\cos \left(a - b\right)}{\sin \left(a - b\right)}$

$\setminus \setminus \setminus = \tan \left(a + b\right) \cdot \cot \left(a - b\right)$

Apr 27, 2017

$T a n \frac{a + b}{T} a n \left(a - b\right)$

#### Explanation:

To solve this, you need to know these formulae:

$S \in \left(x + y\right) = S \in \left(x\right) C o s \left(y\right) + C o s \left(x\right) S \in \left(y\right)$-----equation 1
$C o s \left(x + y\right) = C o s \left(x\right) C o s \left(y\right) - S \in \left(x\right) S \in \left(y\right)$----equation 2

If you replace $y$ with $- y$ in equation 1, we get,
$S \in \left(x - y\right) = S \in \left(x\right) C o s \left(y\right) - C o s \left(x\right) S \in \left(y\right)$

Similarly, replacing $y$ with $- y$ in equation 2, we get,
$C o s \left(x - y\right) = C o s \left(x\right) C o s \left(y\right) + S \in \left(x\right) S \in \left(y\right)$

Now let's move on to the question.
Simplifying all the terms, we get
$\frac{S \in \left(a + b\right)}{C} o s \left(a + b\right) \times \frac{C o s \left(a - b\right)}{S \in \left(a - b\right)}$

$T a n \frac{a + b}{T} a n \left(a - b\right)$ is the answer