How do you prove cos(x-y)/cos(x+y)=(cot x +tan y)/( cot x- tan y)?

Mar 17, 2018

See below...

Explanation:

$\cos \frac{x - y}{\cos} \left(x + y\right)$

$= \frac{\cos x \cdot \cos y + \sin x \cdot \sin y}{\cos x \cdot \cos y - \sin x \cdot \sin y}$

$= \frac{\frac{\cos x \cdot \cos y + \sin x \cdot \sin y}{\sin x \cdot \cos y}}{\frac{\cos x \cdot \cos y - \sin x \cdot \sin y}{\sin x \cdot \cos y}}$

$= \frac{\frac{\cos x \cdot \cos y}{\sin x \cdot \cos y} + \frac{\sin x \cdot \sin y}{\sin x \cdot \cos y}}{\frac{\cos x \cdot \cos y}{\sin x \cdot \cos y} - \frac{\sin x \cdot \sin y}{\sin x \cdot \cos y}}$

$= \frac{\frac{\cos x \cdot \cancel{\cos y}}{\sin x \cdot \cancel{\cos y}} + \frac{\cancel{\sin x} \cdot \sin y}{\cancel{\sin x} \cdot \cos y}}{\frac{\cos x \cdot \cancel{\cos y}}{\sin x \cdot \cancel{\cos y}} - \frac{\cancel{\sin x} \cdot \sin y}{\cancel{\sin x} \cdot \cos y}}$

$= \frac{\cot x + \tan y}{\cot x - \tan y}$

FORMULA reference:- wiki

hope it helps...
Thank you...

Mar 18, 2018

We seek to prove that:

$\cos \frac{x - y}{\cos} \left(x + y\right) \equiv \frac{\cot x + \tan y}{\cot x - \tan y}$

We can the trigonometric identities:

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

Consider the LHS:

$L H S = \cos \frac{x - y}{\cos} \left(x + y\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\cos x \cos y + \sin x \sin y}{\cos x \cos y - \sin x \sin y}$

Now if we multiply both numerator and denominator by $\frac{1}{\sin x \cos y}$ we get:

$L H S = \frac{\cos x \cos y + \sin x \sin y}{\cos x \cos y - \sin x \sin y} \cdot \frac{\frac{1}{\sin x \cos y}}{\frac{1}{\sin x \cos y}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\frac{\cos x \cos y}{\sin x \cos y} + \frac{\sin x \sin y}{\sin x \cos y}}{\frac{\cos x \cos y}{\sin x \cos y} - \frac{\sin x \sin y}{\sin x \cos y}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\frac{\cos x}{\sin x} + \frac{\sin y}{\cos y}}{\frac{\cos x}{\sin x} - \frac{\sin y}{\cos y}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\cot x + \tan y}{\cot x - \tan y} \setminus \setminus \setminus$ QED