\(\frac{1\;-\;cos2x\;+\;sin2x}{1\;+\;cos2x\;+\;sin2x}\) + \(\frac{1\;+\;cosx\;+\;cos2x}{sinx\;+\;sin2x}\)=?
Answer & explanation
Correct answer: option 1
Put x = 30°
⇒ \(\frac{\frac{1}{2} + \frac{\sqrt {3}}{2}}{\frac{3}{2} + \frac{\sqrt {3}}{2}}\) + \(\frac{\frac{3}{2} + \frac{\sqrt {3}}{2}}{\frac{1}{2} + \frac{\sqrt {3}}{2}}\) = \(\frac{1+\sqrt {3}}{3 + \sqrt {3}}\) + \(\frac{3 + \sqrt {3}}{1 + \sqrt {3}}\)
= \(\frac{1+\sqrt {3}}{\sqrt {3}(1+\sqrt {3})}\) + \(\frac{\sqrt {3}(1+\sqrt {3})}{1+\sqrt {3}}\) =\(\frac{1}{\sqrt {3}}\) + 3 = \(\frac{4}{\sqrt {3}}\)
Satisfy from options (1) 2cosec2x
⇒ 2cosec60° = \(\frac{4}{\sqrt {3}}\) (satisfied)