Answer: (a) \(\theta=1.11,\,-2.03,\,-\frac{\pi}{4},\,\frac{3\pi}{4}\); (c) \(\operatorname{cot}x=-\frac{15}{8}\).
Answer: (a) \(\theta=1.11,\,-2.03,\,-\frac{\pi}{4},\,\frac{3\pi}{4}\); (c) \(\operatorname{cot}x=-\frac{15}{8}\).
(a) Use the identity
\(\operatorname{sec}^{2}\theta=1+\tan^{2}\theta.\)
The equation becomes
\(1+\tan^{2}\theta=\tan\theta+3.\)
Rearrange:
\(\tan^{2}\theta-\tan\theta-2=0.\)
Factorise:
\((\tan\theta-2)(\tan\theta+1)=0.\)
So
\(\tan\theta=2 \quad\text{or}\quad \tan\theta=-1.\)
For \(-\pi\lt \theta\lt \pi\), the solutions are
\(\theta=\tan^{-1}2,\quad \tan^{-1}2-\pi,\quad -\frac{\pi}{4},\quad \frac{3\pi}{4}.\)
Thus
\(\theta=1.11,\,-2.03,\,-\frac{\pi}{4},\,\frac{3\pi}{4}.\)
(b) Since
\(\tan\phi=\frac{\sin\phi}{\cos\phi},\)
the left-hand side is
\(\frac{\tan\phi}{\sqrt{1-\cos^{2}\phi}} =\frac{\frac{\sin\phi}{\cos\phi}}{\sqrt{1-\cos^{2}\phi}}.\)
Using \(1-\cos^{2}\phi=\sin^{2}\phi\), this becomes
\(\frac{\frac{\sin\phi}{\cos\phi}}{\sqrt{\sin^{2}\phi}}.\)
Because \(0\lt \phi\lt \frac{\pi}{2}\), \(\sin\phi\gt 0\), so \(\sqrt{\sin^{2}\phi}=\sin\phi\).
Therefore
\(\frac{\frac{\sin\phi}{\cos\phi}}{\sin\phi} =\frac{1}{\cos\phi} =\operatorname{sec}\phi.\)
(c) Since
\(\operatorname{cosec}x=-\frac{17}{8},\)
we have
\(\sin x=-\frac{8}{17}.\)
Also, \(\frac{3\pi}{2}\lt x\lt 2\pi\), so \(x\) is in the fourth quadrant and \(\cos x\) is positive.
Using \(\sin^{2}x+\cos^{2}x=1\),
\(\cos x=\sqrt{1-\left(-\frac{8}{17}\right)^2} =\sqrt{\frac{225}{289}} =\frac{15}{17}.\)
Hence
\(\operatorname{cot}x=\frac{\cos x}{\sin x} =\frac{\frac{15}{17}}{-\frac{8}{17}} =-\frac{15}{8}.\)
