Find the value of \(\frac{cos^235°\;+\;cos^255°}{sin^290°}\) - cot245° + tan260°
Answer & explanation
Correct answer: option 4
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