Math, asked by leha34, 5 months ago

plzzzz solve it..... ​

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Answers

Answered by BrainlyEmpire
34

Question:-

If \sf\:\tan\:\theta\:=\:\dfrac{1}{\sqrt{5}}, find the value of

\displaystyle\sf\:\dfrac{\csc^2\:\theta\:-\:\sec^2\:\theta}{\csc^2\:\theta\:+\:\sec^2\:\theta}

Answer:-

\displaystyle\boxed{\red{\sf\:\dfrac{\csc^2\:\theta\:-\:\sec^2\:\theta}{\csc^2\:\theta\:+\:\sec^2\:\theta}\:=\:\dfrac{2}{3}}}

Step-by-step-explanation:-

We have given that,

\sf\:\tan\:\theta\:=\:\dfrac{1}{\sqrt{5}}

We know that,

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

\displaystyle\implies\sf\:\cot\:\theta\:=\:\dfrac{1}{\dfrac{1}{\sqrt{5}}}

\displaystyle\implies\sf\:\cot\:\theta\:=\:\dfrac{1}{1}\:\times\:\dfrac{\sqrt{5}}{1}

\displaystyle\implies\sf\:\cot\:\theta\:=\:\dfrac{\sqrt{5}}{1}

\displaystyle\implies\boxed{\red{\sf\:\cot\:\theta\:=\:\sqrt{5}}}

Now, we know that,

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

\displaystyle\implies\sf\:\csc^2\:\theta\:=\:1\:+\:(\:\sqrt{5}\:)^2

\displaystyle\implies\sf\:\csc^2\:\theta\:=\:1\:+\:5

\displaystyle\implies\boxed{\red{\sf\:\csc^2\:\theta\:=\:6}}

Now, we know that,

\displaystyle\pink{\sf\:\sec^2\:\theta\:=\:1\:+\:\tan^2\:\theta}\sf\:\:\:-\:-\:-\:[\:Trigonometric\:identity\:]

\displaystyle\implies\sf\:\sec^2\:\theta\:=\:1\:+\:\left(\:\dfrac{1}{\sqrt{5}}\:\right)^2

\displaystyle{\implies\sf\:\sec^2\:\theta\:=\:1\:+\:\left(\:\dfrac{\:(\:1\:)^2}{\;(\:\sqrt{5}\:)^2}\:\right)\:\:\:-\:-\:[\:\because\:\left(\:\dfrac{a}{b}\:\right)^2\:=\:\dfrac{a^2}{b^2}\:]}

\displaystyle\implies\sf\:\sec^2\:\theta\:=\:1\:+\:\dfrac{1}{5}

\displaystyle\implies\sf\:\sec^2\:\theta\:=\:\dfrac{1\:\times\:5\:+\:1}{5}

\displaystyle\implies\sf\:\sec^2\:\theta\:=\:\dfrac{5\:+\:1}{5}

\displaystyle\implies\boxed{\red{\sf\:\sec^2\:\theta\:=\:\dfrac{6}{5}}}

Now,

\displaystyle\sf\:\dfrac{\csc^2\:\theta\:-\:\sec^2\:\theta}{\csc^2\:\theta\:+\:\sec^2\:\theta}

\displaystyle\implies\sf\:\dfrac{6\:-\:\dfrac{6}{5}}{6\:+\:\dfrac{6}{5}}

\displaystyle\implies\sf\:\dfrac{\dfrac{6\:\times\:5\:-\:6}{5}}{\dfrac{6\:\times\:5\:+\:6}{5}}

\displaystyle\implies\sf\:\dfrac{\dfrac{30\:-\:6}{5}}{\dfrac{30\:+\:6}{5}}

\displaystyle\implies\sf\:\dfrac{\dfrac{24}{5}}{\dfrac{36}{5}}

\displaystyle\implies\sf\:\dfrac{24}{\cancel{5}}\:\times\:\dfrac{\cancel{5}}{36}

\displaystyle\implies\sf\:\cancel{\dfrac{24}{36}}

\displaystyle\therefore\boxed{\red{\sf\:\dfrac{\csc^2\:\theta\:-\:\sec^2\:\theta}{\csc^2\:\theta\:+\:\sec^2\:\theta}\:=\:\dfrac{2}{3}}}

Answered by Anonymous
42

Answer:

Answer:-

\displaystyle\boxed{\blue{\sf\:\dfrac{\csc^2\:\theta\:-\:\sec^2\:\theta}{\csc^2\:\theta\:+\:\sec^2\:\theta}\:=\:\dfrac{2}{3}}}

Step-by-step-explanation:-

We have given that,

\sf\:\tan\:\theta\:=\:\dfrac{1}{\sqrt{5}}

We know that,

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

\displaystyle\implies\sf\:\cot\:\theta\:=\:\dfrac{1}{\dfrac{1}{\sqrt{5}}}

\displaystyle\implies\sf\:\cot\:\theta\:=\:\dfrac{1}{1}\:\times\:\dfrac{\sqrt{5}}{1}

\displaystyle\implies\sf\:\cot\:\theta\:=\:\dfrac{\sqrt{5}}{1}

\displaystyle\implies\boxed{\red{\sf\:\cot\:\theta\:=\:\sqrt{5}}}

Now, we know that,

\displaystyle\green{\sf\:\csc^2\:\theta\:=\:1\:+\:\cot^2\:\theta}\sf\:\:\:-\:-\:[\:Trigonometric\:identity\:]

\displaystyle\implies\sf\:\csc^2\:\theta\:=\:1\:+\:(\:\sqrt{5}\:)^2

\displaystyle\implies\sf\:\csc^2\:\theta\:=\:1\:+\:5

\displaystyle\implies\boxed{\orange{\sf\:\csc^2\:\theta\:=\:6}}

Now, we know that,

\displaystyle\pink{\sf\:\sec^2\:\theta\:=\:1\:+\:\tan^2\:\theta}\sf\:\:\:-\:-\:-\:[\:Trigonometric\:identity\:]

\displaystyle\implies\sf\:\sec^2\:\theta\:=\:1\:+\:\left(\:\dfrac{1}{\sqrt{5}}\:\right)^2

\displaystyle{\implies\sf\:\sec^2\:\theta\:=\:1\:+\:\left(\:\dfrac{\:(\:1\:)^2}{\;(\:\sqrt{5}\:)^2}\:\right)\:\:\:-\:-\:[\:\because\:\left(\:\dfrac{a}{b}\:\right)^2\:=\:\dfrac{a^2}{b^2}\:]}

\displaystyle\implies\sf\:\sec^2\:\theta\:=\:1\:+\:\dfrac{1}{5}

\displaystyle\implies\sf\:\sec^2\:\theta\:=\:\dfrac{1\:\times\:5\:+\:1}{5}

\displaystyle\implies\sf\:\sec^2\:\theta\:=\:\dfrac{5\:+\:1}{5}

\displaystyle\implies\boxed{\pink{\sf\:\sec^2\:\theta\:=\:\dfrac{6}{5}}}

Now,

\displaystyle\sf\:\dfrac{\csc^2\:\theta\:-\:\sec^2\:\theta}{\csc^2\:\theta\:+\:\sec^2\:\theta}

\displaystyle\implies\sf\:\dfrac{6\:-\:\dfrac{6}{5}}{6\:+\:\dfrac{6}{5}}

\displaystyle\implies\sf\:\dfrac{\dfrac{6\:\times\:5\:-\:6}{5}}{\dfrac{6\:\times\:5\:+\:6}{5}}

\displaystyle\implies\sf\:\dfrac{\dfrac{30\:-\:6}{5}}{\dfrac{30\:+\:6}{5}}

\displaystyle\implies\sf\:\dfrac{\dfrac{24}{5}}{\dfrac{36}{5}}

\displaystyle\implies\sf\:\dfrac{24}{\cancel{5}}\:\times\:\dfrac{\cancel{5}}{36}

\displaystyle\implies\sf\:\cancel{\dfrac{24}{36}}

\displaystyle\therefore\boxed{\orange{\sf\:\dfrac{\csc^2\:\theta\:-\:\sec^2\:\theta}{\csc^2\:\theta\:+\:\sec^2\:\theta}\:=\:\dfrac{2}{3}}}

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