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Answer 1

$\bf \huge \underline{ \orange{Required \: Answer : }}</p><p>$

Given secΘ = 10/9 , calculate all other trigonometric ratios.

$\bf \huge \underline{ \orange{answer : }}</p><p>$

$\begin{gathered}\sf{sec \: \theta = \frac{hyptenuse}{adjecent} = \frac{10}{9} } \\ \\\end{gathered}</p><p>$

$\bf{we \: will \: find \:the \: oppsite \: side }$

$\begin{gathered}\to \tt{according \: to \: the \: attached \: diagram } \\\end{gathered}$

AB² + BC²=AC²

➜ AB² = AC² - BC²

➜ AB² = 10² - 9²

➜ AB² = 100 - 81

➜ AB² = 27

∴ AB = √27

$\sf \small{all \: other \: trigonometric \: ratios \:are \: as \: follows : }$

$\begin{gathered}\to \tt{sin \theta = \frac{opposite}{hypotenuse} = \frac{ \sqrt{27} }{10} } \\ \\ \to \tt{cos \ theta = \frac{adjecent}{hypotenuse} = \frac{9}{10} } \\ \\ \to \tt{tan \theta = \frac{opposite}{adjecent} = \frac{ \sqrt{27} }{9} } \\ \\ \to \tt{cot \theta = \frac{adjecent}{opposite} = \frac{9}{ \sqrt{27} } } \\ \\ \to \tt{cosec \theta = \frac{hypotenuse}{opposite} = \frac{10}{ \sqrt{27} } }\end{gathered}</p><p>$

Answer 2

Given secΘ = 10/9 , calculate all other trigonometric ratios.

$\begin{gathered}\begin{gathered}\sf{sec \: \theta = \frac{hyptenuse}{adjecent} = \frac{10}{9} } \\ \\\end{gathered} < /p > < p > \end{gathered}$

$\bf{we \: will \: find \:the \: oppsite \: side }$

$\begin{gathered}\begin{gathered}\to \tt{according \: to \: the \: attached \: diagram } \\\end{gathered} \end{gathered}$

AB² + BC²=AC²

➜ AB² = AC² - BC²

➜ AB² = 10² - 9²

➜ AB² = 100 - 81

➜ AB² = 27

∴ AB = √27

$\sf \small{all \: other \: trigonometric \: ratios \:are \: as \: follows : }$

$\begin{gathered}\begin{gathered}\to \tt{sin \theta = \frac{opposite}{hypotenuse} = \frac{ \sqrt{27} }{10} } \\ \\ \to \tt{cos \ theta = \frac{adjecent}{hypotenuse} = \frac{9}{10} } \\ \\ \to \tt{tan \theta = \frac{opposite}{adjecent} = \frac{ \sqrt{27} }{9} } \\ \\ \to \tt{cot \theta = \frac{adjecent}{opposite} = \frac{9}{ \sqrt{27} } } \\ \\ \to \tt{cosec \theta = \frac{hypotenuse}{opposite} = \frac{10}{ \sqrt{27} } }\end{gathered} < /p > < p > \end{gathered}$