Given cot A + tan A = 5, find cot^2 A + tan^2 A

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Given cot A + tan A = 5, find cot^2 A + tan^2 A

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Katherine 2 months 2021-11-23T02:51:49+00:00 2 Answers 0 views 0

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    0
    2021-11-23T02:53:34+00:00

    \LARGE\star\boxed{\mathfrak\pink{\underline{\underline{Answer}}}}\star\\\\\\

    \Large\textbf{$cot^2\:A\:+\:tan^2\:A\:=\:23$}\\\\\\

    \LARGE\star\star\boxed{\mathbb\red{\underline{\underline{GIVEN}}}}\star\star\\\\\\

    \Large\odot\:\:\textbf{cot\:A\:+\:tan\:A\:=\:5}\\\\\\

    \LARGE\star\star\star\boxed{\mathbb\green{\underline{\underline{TO\:FIND}}}}\star\star\star\\\\\\

    \Large\odot\:\:\textbf{Value\:of\:$cot^2\:A\:+\:tan^2\:A$}\\\\\\

    \LARGE\star\star\star\star\boxed{\mathcal\red{\underline{\underline{Explanation}}}}\star\star\star\star\\\\\\

    \Large\texttt{Squaring LHS and RHS}\\\\\\

    \Large\texttt{Using $(a+b)^2\:=\:a^2+b^2+2ab$}\\\\\\

    \large\Longrightarrow\textsf{$cot^2\:A\:+\:tan^2\:A\:+2\:cot\:A\:tan\:A\:=\:5^2$}\\\\\\

    \Large\texttt{We know that cot A=$\dfrac{1}{tan\:A}$}\\\\\\

    \large\Longrightarrow\textsf{$cos^2\:A\:+\:tan^2\:A\:+\:2\:\dfrac{1}{tan\:A}\times\:tan\:A\:=\:25$}\\\\

    \large\Longrightarrow\textsf{$cos^2\:A\:+\:tan^2\:A\:+2\:=\:25$}\\\\\\

    \Large\textsf{Transfering 2 from LHS to RHS}\\\\\\

    \large\Longrightarrow\textsf{$cos^2\:A\:+\:tan^2\:A\:=\:23$}\\\\

    \Large\therefore\textbf{The\:Answer\:is\:}\\

    \Large\textbf{$cos^2\:A\:+\:tan^2\:A\:=\:23$}\\\\

    \LARGE\mathcal{\underline{NOTE}}\\\\

    \large\odot\:\:\texttt{In such types of questions}\\

    \large\texttt{try to solve the question}\\

    \large\texttt{by either changing every}\\

    \large\texttt{term into a single term or }\\

    \large\texttt{using some of the trigonometric}\\

    \large\texttt{identities or relations.}\\\\\\

    \Large\mathcal{\underline{Trigonometric\:Identities}}\\\\\\

    \large\odot\:\:\texttt{$sin^2\:A\:+\:cos^2\:A\:=\:1$}\\\\

    \large\odot\:\:\texttt{$cosec^2\:A\:-\:cot^2\:A\:=\:1$}\\\\

    \large\odot\:\:\texttt{$sec^2\:A\:-\:tan^2\:A\:=\:1$}\\\\\\

    \Large\mathcal{\underline{Trigonometric\:Relations}}\\\\\\

    \large\texttt{sin A = $\dfrac{1}{cosec\:A}$}\\

    \large\texttt{or cosec A = $\dfrac{1}{sin\:A}$}\\\\

    \large\texttt{cos A = $\dfrac{1}{sec\:A}$}\\

    \large\texttt{or sec A = $\dfrac{1}{cos\:A}$}\\\\

    \large\texttt{tan A = $\dfrac{1}{cot\:A}$}\\

    \large\texttt{or cot A = $\dfrac{1}{tan\:A}$}\\\\

    \large\texttt{tan A = $\dfrac{sin\:A}{cos\:A}$}\\\\

    \large\texttt{cot A = $\dfrac{cos\:A}{sin\:A}$}\\\\\\

    ▬▬▬▬▬ஜ۩۞۩ஜ▬▬▬▬▬▬

    0
    2021-11-23T02:53:38+00:00

    Answer:

    tanA + cotA = 5

    (tanA +cotA)2 = tanA2 + cotA2 + 2(tanA * cotA)

    25 = tanA2 + cot2A + 2

    tanA *cotA = 1

    tan2A + cot2A = 23

    Step-by-step explanation:

    hope it helps ✌️✌️

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