What is the value of [(cos 5A + cos 3A)] ÷ [( sin 5A - sin 3A)] ?

Here,

(cos A + cos B = 2 cos (\(\frac{A+B}{2}\)) cos (\(\frac{A-B}{2}\)) and

(sin A - sin B = 2 cos (\(\frac{A+B}{2}\)) sin (\(\frac{A-B}{2}\))

Here using these formulas:

⇒ \(\frac{cos 5A + cos 3A}{(sin 5A + sin 3A)}\)

= \(\frac{2cos\;\frac{(5A\;+\;3A)}{2}\;cos\;\frac{(5A\;-\;3A)}{2}}{2cos\;\frac{(5A\;+\;3A)}{2}\;sin\;\frac{(5A\;-\;3A)}{2}}\)

= \(\frac{2cos4A\;cosA}{2cos 4A\;sinA}\)

= cot A