# How does one evaluate the identity and prove it to be correct?

## cos(x-y)/cos(x)cos(y)=1+tan(x)tan(y)

Apr 5, 2018

Formula reference...

Hope it helps...
Thank you....

Apr 5, 2018

See below.

#### Explanation:

Given identity:
$\cos \frac{x - y}{\cos \left(x\right) \cos \left(y\right)} = 1 + \tan \left(x\right) \tan \left(y\right)$

Since we don't have values for $x$ and $y$, we can't actually evaluate this identity, however we can prove the identity:

In order to prove the identity, we must either manipulate the left hand side to be equal to the right hand side or vice versa or manipulate both sides to be equal at some middle point.

Consider the sum/difference formulas for cosine:
$\cos \left(a + b\right) = \cos \left(a\right) \cos \left(b\right) - \sin \left(a\right) \sin \left(b\right)$
$\cos \left(a - b\right) = \cos \left(a\right) \cos \left(b\right) + \sin \left(a\right) \sin \left(b\right)$

We will start from the left side and move to the right side:
Noting that $\cos \left(a - b\right) = \cos \left(a\right) \cos \left(b\right) + \sin \left(a\right) \sin \left(b\right)$:
$L H S = \frac{\cos \left(x\right) \cos \left(y\right) + \sin \left(x\right) \sin \left(y\right)}{\cos \left(x\right) \cos \left(y\right)}$

Separating the above fraction into two fractions, we have:
$L H S = \frac{\cos \left(x\right) \cos \left(y\right)}{\cos \left(x\right) \cos \left(y\right)} + \frac{\sin \left(x\right) \sin \left(y\right)}{\cos \left(x\right) \cos \left(y\right)}$

Simplifying and noting that $\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$, we have:
$L H S = 1 + \tan \left(x\right) \tan \left(y\right) = R H S$
as required

Therefore, $\cos \frac{x - y}{\cos \left(x\right) \cos \left(y\right)} = 1 + \tan \left(x\right) \tan \left(y\right)$