Show that $$\(\frac{\tan \theta}{1+\cos \theta}+\frac{\tan \theta}{1-\cos \theta} \equiv \frac{2}{\sin \theta \cos \theta}\)$$. Hence solve the equation $$\(\frac{\tan \theta}{1+\cos \theta}+\frac{\tan \theta}{1-\cos \theta}=\frac{6}{\tan \theta}\)$$ for $$\(0^{\circ}< \theta< 180^{\circ}\)$$. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_s20_qp_13 Year：2020 Question No：7(b)
Answer:
\(\frac{2}{\sin \theta \cos \theta}=\frac{6 \cos \theta}{\sin \theta}\)
\(\cos ^{2} \theta=\frac{1}{3} \rightarrow \cos \theta=(\pm) 0.5774\)
\(54.7^{\circ}, 125.3^{\circ}\)
(FT for \(180^{\circ}-1\) st solution)
\(\cos ^{2} \theta=\frac{1}{3} \rightarrow \cos \theta=(\pm) 0.5774\)
\(54.7^{\circ}, 125.3^{\circ}\)
(FT for \(180^{\circ}-1\) st solution)
Knowledge points:
1.5.5 find all the solutions of simple trigonometrical equations lying in a specified interval (general forms of solution are not included).
Solution:
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