1.

Cosec A =√10 find all the trigonometric ratio​

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ANSWER:

\displaystyle{\boxed{\red{\sf\:\sin\:A\:=\:\dfrac{1}{\sqrt{10}}}}}

\displaystyle{\boxed{\pink{\sf\:\cos\:A\:=\:\dfrac{3}{\sqrt{10}}}}}

\displaystyle{\boxed{\green{\sf\:\tan\:A\:=\:\dfrac{1}{3}}}}

\displaystyle{\boxed{\blue{\sf\:\sec\:A\:=\:\dfrac{\sqrt{10}}{3}}}}

\displaystyle{\boxed{\red{\sf\:\cot\:A\:=\:3}}}

Step-by-step-explanation:

We have GIVEN that,

\displaystyle\sf\:cosec\:A\:=\:\sqrt{10}

We have to FIND all the REMAINING TRIGONOMETRIC ratios.

Now,

\displaystyle\sf\:cosec\:A\:=\:\sqrt{10}

\displaystyle{\pink{\sf\:\sin\:A\:=\:\dfrac{1}{cosec\:A}}\sf\:\:\:-\:-\:-\:[\:Trigonometric\:identity\:]}

\displaystyle{\implies\boxed{\red{\sf\:\sin\:A\:=\:\dfrac{1}{\sqrt{10}}}}}

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Now,

\displaystyle{\pink{\sf\:1\:+\:\cot^2\:A\:=\:cosec^2\:A}\sf\:\:\:-\:-\:[\:Trigonometric\:identity\:]}

\displaystyle{\implies\sf\:1\:+\:\cot^2\:A\:=\:\left(\:\sqrt{10}\:\right)^2}

\displaystyle{\implies\sf\:1\:+\:\cot^2\:A\:=\:10}

\displaystyle{\implies\sf\:\cot^2\:A\:=\:10\:-\:1}

\displaystyle{\implies\sf\:\cot^2\:A\:=\:9}

\displaystyle{\implies\sf\:\cot\:A\:=\:\sqrt{9}}

\displaystyle{\implies\boxed{\pink{\sf\:\cot\:A\:=\:3}}}

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Now,

\displaystyle{\pink{\sf\:\tan\:A\:=\:\dfrac{1}{\cot\:A}}\sf\:\:\:-\:-\:-\:[\:Trigonometric\:identity\:]}

\displaystyle{\implies\boxed{\green{\sf\:\tan\:A\:=\:\dfrac{1}{3}}}}

─────────────────────

Now,

\displaystyle{\pink{\sf\:\tan\:A\:=\:\dfrac{\sin\:A}{\cos\:A}}\sf\:\:\:-\:-\:[\:Trigonometric\:identity\:]}

\displaystyle{\implies\sf\:\dfrac{1}{3}\:=\:\dfrac{\dfrac{1}{\sqrt{10}}}{\cos\:A}}

\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{\dfrac{1}{\sqrt{10}}}{\dfrac{1}{3}}}

\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{1}{\sqrt{10}}\:\times\:\dfrac{3}{1}}

\displaystyle{\implies\boxed{\blue{\sf\:\cos\:A\:=\:\dfrac{3}{\sqrt{10}}}}}

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Now,

\displaystyle{\pink{\sf\:\sec\:A\:=\:\dfrac{1}{\cos\:A}}\sf\:\:\:-\:-\:[\:Trigonometric\:identity\:]}

\displaystyle{\implies\sf\:\sec\:A\:=\:\dfrac{1}{\dfrac{3}{\sqrt{10}}}}

\displaystyle{\implies\sf\:\sec\:A\:=\:1\:\times\:\dfrac{\sqrt{10}}{3}}

\displaystyle{\implies\boxed{\red{\sf\:\sec\:A\:=\:\dfrac{\sqrt{10}}{3}}}}


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