Correct option is: A) \( \frac{3}{5}\)

We have cosec \(\theta\) + cot \(\theta\) = 2

= \(\frac {1}{sin \theta} + \frac {cos \theta}{sin \theta} =2\)

= 1 + cos \(\theta\) = 2 sin \(\theta\)

= \((1 + cos \theta )^2 = 4\, sin^2\theta\) (By squaring both sides)

= 1 + \(cos^2\theta + 2\, cos \theta = 4 (1- cos^2\theta)\) (\(\because\) \(sin^2\theta = 1- cos^2\theta\))

= \(5 \,cos ^2\theta + 2\, cos \theta - 3 = 0\)

= \(5 \,cos ^2\theta + 5 \,cos\theta - 3 \, cos\theta - 3 = 0\)

= \(5 \,cos\theta (cos\theta +1) - 3 (cos\theta +1 ) = 0\)

= (5 cos \(\theta\) - 3 ) (cos \(\theta\) + 1) = 0

= 5 cos \(\theta\) - 3 = 0 or cos \(\theta\) + 1 = 0

= cos \(\theta\) = \(\frac 35\) or cos \(\theta\) = -1

cos \(\theta\) \(\neq\) -1

\(\because\) If cos \(\theta\) = -1 then sin \(\theta\) = \(\sqrt {1-cos^2\theta} = \sqrt {1-(-1)^2} = \sqrt {1-1} = 0\)

Then cosec \(\theta\) = \(\frac 1{sin \theta} = \frac 10 = \infty\) (Not defined)

\(\therefore\) cos \(\theta\) = \(\frac 35\)

Correct option is: A) \(\frac{3}{5}\)