Find the value of \(\frac{cos^235°\;+\;cos^255°}{sin^290°}\) - cot245° + tan260°

2 1 0 3

3

Formula → [cos2A + cos2B = 1, when A + B = 90°] Here, (35° + 55° = 90°), therefore, ⇒ \(\frac{cos^235°\;+\;cos^255°}{sin^290°}\) - cot245° + tan260° = \(\frac{1}{1}\) - 1 + (\(\sqrt {3}\))2 = 1 - 1 + 3 = 3