Find the value of \(\frac{cos^225°\;+\;cos^265°}{sin^290°}\) - tan245° + cot230°

0 1 2 3

3

Formula → [cos2A + cos2B = 1, when A + B = 90°] Here, (25° + 65° = 90°), therefore, ⇒ \(\frac{cos^225°\;+\;cos^265°}{sin^290°}\) - tan245° + cot230° = \(\frac{1}{1}\) - 1 + (\(\sqrt {3}\))2 = 1 - 1 + 3 = 3