Question

2. Establish the identity (prove the two sides are equal)

cos(A-B)/ (sin A sin B) = cot A + tan B​

Answer 1

Answer:

See below!

Step-by-step explanation:

To prove the given trigonometric identity $\sf \dfrac{\cos(A-B)}{\sin A \sin B} = \cot A + \tan B$, we'll start by expressing each side of the equation in terms of basic trigonometric functions.

Right-hand side (RHS):

$\sf \cot A + \tan B = \dfrac{\cos A}{\sin A} + \dfrac{\sin B}{\cos B}$

Now, let's find a common denominator for the fractions:

$\sf \dfrac{\cos A \cos B + \sin A \sin B}{\sin A \cos B}$

Use the angle subtraction formula $\sf ( \cos(A - B) = \cos A \cos B + \sin A \sin B)$:

$\sf \dfrac{\cos(A - B)}{\sin A \sin B}$

Now, the right-hand side (RHS) is equal to the left-hand side (LHS), and the identity is established:

$\sf \dfrac{\cos(A - B)}{\sin A \sin B} = \cot A + \tan B$

Therefore, the given trigonometric identity is proven to be true.