Question

Let 0 be an angle in quadrant iv such that

Answer 1

`\begin{gathered} \tan \theta=-(8)/(15) \\ \cos ^{}\theta=(15)/(17) \end{gathered}`

Step-by-step explanation

Given:

`\begin{gathered} \theta\text{ is in quadrant VI,} \\ \csc \theta=-(17)/(8) \end{gathered}`

Using trigonometric identity, To find the exact value of tan θ

`1+\cot ^2\theta=\csc ^2\theta`
`\begin{gathered} 1+\cot ^2\theta=(-(17)/(8))^2 \\ 1+\cot ^2\theta=(289)/(64) \\ \cot ^2\theta=(289)/(64)-1 \\ \cot ^2\theta=(289-64)/(64) \\ \cot ^2\theta=(225)/(64) \\ \cot ^{}\theta=\pm\sqrt[]{(225)/(64)} \\ \cot \theta=\pm(15)/(8) \\ \text{But cot }\theta=(1)/(\tan \theta) \\ \Rightarrow\text{tan }\theta=\pm(8)/(15) \\ \text{Since }\theta\text{ is in quadrant VI, then tan }\theta\text{ is negative} \\ \tan \theta=-(8)/(15) \end{gathered}`

To find the exact value of cos θ

`\begin{gathered} \text{Using trigonometric identuty} \\ \cos ^2\theta=(1)/(1+\tan ^2\theta) \\ \cos ^2\theta=(1)/(1+((64)/(225))) \\ \cos ^2\theta=(1)/((289)/(225)) \\ \cos ^2\theta=(225)/(289) \\ \cos ^{}\theta=\pm\sqrt[]{(225)/(289)} \\ \cos ^{}\theta=\pm(15)/(17) \\ \text{Since }\theta\text{ is in quadrant VI, cos }\theta\text{ is positve} \\ \therefore\cos ^{}\theta=(15)/(17) \end{gathered}`

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