Answer: \(\phi=-405^\circ,\ -135^\circ,\ 135^\circ,\ 405^\circ\).
Start by writing the trigonometric functions in terms of \(\sin\), \(\cos\), \(\tan\), \(\operatorname{sec}\), or \(\operatorname{cosec}\) as appropriate, then simplify algebraically.
(a) Start with the left-hand side:
\(\frac{\operatorname{cot}\theta+\tan\theta}{\operatorname{sec}\theta}.\)
Use
\(\operatorname{cot}\theta=\frac{\cos\theta}{\sin\theta},\qquad \tan\theta=\frac{\sin\theta}{\cos\theta},\qquad \operatorname{sec}\theta=\frac1{\cos\theta}.\)
Then
\(\operatorname{cot}\theta+\tan\theta =\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta} =\frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta} =\frac1{\sin\theta\cos\theta}.\)
Therefore
\(\frac{\operatorname{cot}\theta+\tan\theta}{\operatorname{sec}\theta} =\frac{\frac1{\sin\theta\cos\theta}}{\frac1{\cos\theta}} =\frac1{\sin\theta} =\operatorname{cosec}\theta.\)
(b) Let
\(u=\frac{\phi}{3}.\)
Using the identity from part (a), the equation becomes
\(\operatorname{cosec}^2u=2.\)
So
\(\sin^2u=\frac12.\)
Hence
\(\sin u=\pm\frac1{\sqrt2}.\)
Since
\(-540^\circ\lt\phi\lt540^\circ,\)
we have
\(-180^\circ\lt u\lt180^\circ.\)
In this interval,
\(u=-135^\circ,\ -45^\circ,\ 45^\circ,\ 135^\circ.\)
Multiplying by 3 gives
\(\phi=-405^\circ,\ -135^\circ,\ 135^\circ,\ 405^\circ.\)
