Answer: \(\displaystyle \phi=-90^\circ,-67.5^\circ,-22.5^\circ,0^\circ,22.5^\circ,67.5^\circ,90^\circ\).
Answer: \(\displaystyle \phi=-90^\circ,-67.5^\circ,-22.5^\circ,0^\circ,22.5^\circ,67.5^\circ,90^\circ\).
(a) Start with the left-hand side:
\(\frac{1}{\operatorname{cosec}\theta-1}+ \frac{1}{\operatorname{cosec}\theta+1}.\)
Use a common denominator:
\(\frac{\operatorname{cosec}\theta+1+\operatorname{cosec}\theta-1} {(\operatorname{cosec}\theta-1)(\operatorname{cosec}\theta+1)}.\)
This simplifies to
\(\frac{2\operatorname{cosec}\theta}{\operatorname{cosec}^{2}\theta-1}.\)
Since
\(\operatorname{cosec}^{2}\theta-1=\operatorname{cot}^{2}\theta,\)
we get
\(\frac{2\operatorname{cosec}\theta}{\operatorname{cot}^{2}\theta}.\)
Now write this in terms of sine and cosine:
\(\frac{2(1/\sin\theta)}{\cos^{2}\theta/\sin^{2}\theta} =2\sin\theta\cdot\frac{1}{\cos^{2}\theta}.\)
Therefore
\(\frac{1}{\operatorname{cosec}\theta-1}+ \frac{1}{\operatorname{cosec}\theta+1} =2\sin\theta\,\operatorname{sec}^{2}\theta.\)
(b) Using part (a) with \(\theta=2\phi\), the equation becomes
\(2\sin2\phi\,\operatorname{sec}^{2}2\phi=4\sin2\phi.\)
Bring all terms to one side:
\(2\sin2\phi(\operatorname{sec}^{2}2\phi-2)=0.\)
So either
\(\sin2\phi=0\)
or
\(\operatorname{sec}^{2}2\phi=2.\)
From \(\sin2\phi=0\), with \(-90^\circ\leq\phi\leq90^\circ\),
\(\phi=-90^\circ,\ 0^\circ,\ 90^\circ.\)
From \(\operatorname{sec}^{2}2\phi=2\),
\(\cos^{2}2\phi=\frac12.\)
Thus
\(2\phi=\pm45^\circ,\ \pm135^\circ.\)
Hence
\(\phi=\pm22.5^\circ,\ \pm67.5^\circ.\)
Combining these,
\(\phi=-90^\circ,-67.5^\circ,-22.5^\circ,0^\circ,22.5^\circ,67.5^\circ,90^\circ.\)
