Answer: (b) \(y=-\frac{\pi}{12},\ -\frac{5\pi}{12},\ -\frac{13\pi}{12},\ -\frac{17\pi}{12}\).

We start with the main method. Rewrite everything in terms of \(\sin x\) and \(\cos x\), then simplify using standard identities.

(a) Start with the left-hand side:

\(\displaystyle \frac{1}{\sec x-\csc x}+\frac{1}{\sec x+\csc x}\).

Use a common denominator:

\(\displaystyle \frac{\sec x+\csc x+\sec x-\csc x}{(\sec x-\csc x)(\sec x+\csc x)}\).

The numerator simplifies to \(2\sec x\), and the denominator is \(\sec^2x-\csc^2x\). Therefore

\(\displaystyle \frac{1}{\sec x-\csc x}+\frac{1}{\sec x+\csc x}\)

\(\displaystyle =\frac{2\sec x}{\sec^2x-\csc^2x}\).

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

\(\displaystyle \frac{2\sec x}{\sec^2x-\csc^2x}\)

\(\displaystyle =\frac{\frac{2}{\cos x}}{\frac{1}{\cos^2x}-\frac{1}{\sin^2x}}\).

The denominator is

\(\displaystyle \frac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}\).

So

\(\displaystyle \frac{\frac{2}{\cos x}}{\frac{1}{\cos^2x}-\frac{1}{\sin^2x}}\)

\(\displaystyle =\frac{2\sin^2x\cos x}{\sin^2x-\cos^2x}\).

Now simplify the right-hand side of the required identity:

\(\displaystyle \frac{2\cos x}{1-\cot^2x}\)

\(\displaystyle =\frac{2\cos x}{1-\frac{\cos^2x}{\sin^2x}}\).

This becomes

\(\displaystyle \frac{2\cos x}{\frac{\sin^2x-\cos^2x}{\sin^2x}}\)

\(\displaystyle =\frac{2\sin^2x\cos x}{\sin^2x-\cos^2x}\).

This is the same expression as the left-hand side, so the identity is proved.

(b) From

\(\displaystyle 3\tan^2\left(y+\frac{\pi}{4}\right)=1\),

we get

\(\displaystyle \tan^2\left(y+\frac{\pi}{4}\right)=\frac{1}{3}\).

Therefore

\(\displaystyle \tan\left(y+\frac{\pi}{4}\right)=\pm\frac{1}{\sqrt3}\).

Let

\(\displaystyle \theta=y+\frac{\pi}{4}\).

Since \(-2\pi

\(\displaystyle -\frac{7\pi}{4}<\theta<\frac{\pi}{4}\).

In this interval, the solutions for \(\theta\) are

\(\displaystyle \theta=-\frac{7\pi}{6},\ -\frac{5\pi}{6},\ -\frac{\pi}{6},\ \frac{\pi}{6}\).

Subtract \(\frac{\pi}{4}\) from each value:

\(\displaystyle y=-\frac{17\pi}{12},\ -\frac{13\pi}{12},\ -\frac{5\pi}{12},\ -\frac{\pi}{12}\).

So the solutions are

\(\displaystyle y=-\frac{\pi}{12},\ -\frac{5\pi}{12},\ -\frac{13\pi}{12},\ -\frac{17\pi}{12}\).

This completes the solution and gives the required result.