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Trigonometry, 2ADV T2 EQ-Bank 2 MC

Given  \(\tan \theta=\cfrac{1}{3}\)  and  \(0°<\theta<90°\),

find the value of  \(\dfrac{1-\sin (180+\theta)}{\cos (90-\theta)}\).

  1. \(\sqrt{10}+1\)
  2. \(1\)
  3. \(\sqrt{10}-1\)
  4. \(\dfrac{\sqrt{10}}{2}\)
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\(\Rightarrow A\)

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\(\text {Since} \ \ \tan \theta=\dfrac{1}{3}:\)
 

\(\dfrac{1-\sin (180+\theta)}{\cos (90-\theta)}\) \(=\dfrac{1+\sin \theta}{\sin \theta}\)
  \(=\dfrac{1+\frac{1}{\sqrt{10}}}{\frac{1}{\sqrt{10}}} \times \dfrac{\sqrt{10}}{\sqrt{10}}\)
  \(=\sqrt{10}+1\)

 

\(\Rightarrow A\)

Trigonometry, 2ADV T2 EQ-Bank 2

Express  \(3 \operatorname{cosec}(180+x)+5 \cos (90-x)\)  as a single fraction in terms of \(\sin x\), given all angles are measured in degrees.   (3 marks)

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\(\dfrac{-3+5 \sin ^2 x}{\sin x}\)

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\(3 \operatorname{cosec}(180+x)+5 \cos (90-x)\)

\(=\dfrac{3}{\sin \left(180^{\circ}+x\right)}+5 \sin x\)

\(=\dfrac{3}{-\sin x}+5 \sin x\)

\(=\dfrac{-3}{\sin x}+\dfrac{5 \sin ^2 x}{\sin x}\)

\(=\dfrac{-3+5 \sin ^2 x}{\sin x}\)