Answer: (a) \((3\sqrt5,55)\), \((-2\sqrt5,-20)\); (b) \(14+\frac{15\sqrt3}{2}\).

Answer: (a) \((3\sqrt5,55)\), \((-2\sqrt5,-20)\); (b) \(14+\frac{15\sqrt3}{2}\).

(a) Equate the curve and the line:

\(x^2+2\sqrt5x-20=3\sqrt5x+10.\)

So

\(x^2-\sqrt5x-30=0.\)

Using the quadratic formula,

\(x=\frac{\sqrt5\pm\sqrt{125}}2=\frac{\sqrt5\pm5\sqrt5}{2}.\)

Hence \(x=3\sqrt5\) or \(x=-2\sqrt5\). Substituting into \(y=3\sqrt5x+10\) gives \(y=55\) and \(y=-20\). Therefore the points are

\((3\sqrt5,55)\quad\text{and}\quad(-2\sqrt5,-20).\)

(b) Use

\(\operatorname{cosec}^{2}\theta=1+\operatorname{cot}^2\theta.\)

Since

\(\operatorname{cot}\theta=\frac{2+\sqrt3}{\sqrt3-1},\)

rationalise:

\(\operatorname{cot}\theta=\frac{(2+\sqrt3)(\sqrt3+1)}{2}=\frac{5+3\sqrt3}{2}.\)

So

\(\operatorname{cosec}^{2}\theta=1+\left(\frac{5+3\sqrt3}{2}\right)^2.\)

Now \((5+3\sqrt3)^2=52+30\sqrt3\), hence

\(\operatorname{cosec}^{2}\theta=1+\frac{52+30\sqrt3}{4}=14+\frac{15\sqrt3}{2}.\)