Answer: \(\sec30^\circ=\frac{2}{\sqrt3}\), and \(\dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}=2\operatorname{cosec}x\cot x\).

(a) In the equilateral triangle, each side is \(a\). Since \(M\) is the midpoint of \(AC\),

\(AM=\dfrac a2.\)

The line \(BM\) is perpendicular to \(AC\), so triangle \(ABM\) is right-angled. The angle at \(B\) is \(30^\circ\).

Using Pythagoras,

\(BM=\sqrt{a^2-\left(\dfrac a2\right)^2}=\sqrt{\dfrac{3a^2}{4}}=\dfrac{\sqrt3}{2}a.\)

For the angle \(30^\circ\),

\(\cos30^\circ=\dfrac{BM}{AB}=\dfrac{\frac{\sqrt3}{2}a}{a}=\dfrac{\sqrt3}{2}.\)

Therefore

\(\sec30^\circ=\dfrac1{\cos30^\circ}=\dfrac2{\sqrt3}.\)

(b) Start with

\(\dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}.\)

Use a common denominator:

\(\dfrac{(\sec x+1)+(\sec x-1)}{(\sec x-1)(\sec x+1)}.\)

The numerator simplifies to \(2\sec x\), and the denominator is

\(\sec^2x-1.\)

Since \(\sec^2x-1=\tan^2x\), the expression becomes

\(\dfrac{2\sec x}{\tan^2x}.\)

Now write everything in terms of \(\sin x\) and \(\cos x\):

\(\dfrac{2\cdot\frac1{\cos x}}{\frac{\sin^2x}{\cos^2x}} =\dfrac{2}{\cos x}\cdot\dfrac{\cos^2x}{\sin^2x} =\dfrac{2\cos x}{\sin^2x}.\)

Finally,

\(\dfrac{2\cos x}{\sin^2x}=2\left(\dfrac1{\sin x}\right)\left(\dfrac{\cos x}{\sin x}\right)=2\operatorname{cosec}x\cot x.\)

This proves the identity.

Answer: \(y=3\left(\left(\frac{3}{2-x}\right)^2-1\right)\).

We are given

\(x=2-3\sin\theta.\)

Rearrange this to express \(\sin\theta\) in terms of \(x\):

\(3\sin\theta=2-x,\)

so

\(\sin\theta=\dfrac{2-x}{3}.\)

Also,

\(y=3\cot^2\theta.\)

Using

\(\cot^2\theta=\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{1-\sin^2\theta}{\sin^2\theta},\)

we get

\(y=3\left(\dfrac{1-\sin^2\theta}{\sin^2\theta}\right).\)

Substitute \(\sin\theta=\dfrac{2-x}{3}\):

\(y=3\left(\dfrac{1-\left(\dfrac{2-x}{3}\right)^2}{\left(\dfrac{2-x}{3}\right)^2}\right).\)

This is the same as

\(y=3\left(\dfrac{1}{\left(\dfrac{2-x}{3}\right)^2}-1\right).\)

Therefore

\(y=3\left(\left(\dfrac{3}{2-x}\right)^2-1\right).\)