Question

$\sf{if \: a \: b \: c \: and \: d \: are \: in \: proportion \: prove \: that} \\ \sf{abcd( \: \frac{1}{a}^{2} +\frac{1}{b}^{2}\frac{1}{c}^{2} \: \frac{1}{d}^{2})= {a}^{2} {b}^{2} {c}^{2}}$
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Answer 1

Answer:

$\sf \: \red{Solution:-}$

$\sf \pink{Given:-}$

$\begin{gathered}\\\tt \longrightarrow\dfrac{a}{b} = \dfrac{c}{d} = k(const)\\ \end{gathered}$

$\begin{gathered}\\\tt \longrightarrow \: a= bk \: \: and \: \: c = dk\\ \end{gathered}$

$\begin{gathered}\\\tt \: \: \longrightarrow \: \: abcd \left( \dfrac{1}{a^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{c^{2}}+ \dfrac{1}{d^{2}} \right)\\ \end{gathered}$

$\green{ \sf \: ✠Now \: put \: the \: values –}$

$\begin{gathered}\\\tt \: \: \longrightarrow \: \: (bk)b(dk)d \left( \dfrac{1}{(bk)^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{(dk)^{2}}+ \dfrac{1}{d^{2}} \right)\\ \end{gathered}$

$\begin{gathered}\\\tt \: \: \longrightarrow \: \: b^{2} {d}^{2} k^{2} \left( \dfrac{1}{(bk)^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{(dk)^{2}}+ \dfrac{1}{d^{2}} \right)\\ \end{gathered}$

$\begin{gathered}\\\tt \: \: \longrightarrow \: \: b^{2} {d}^{2} k^{2} \left[\dfrac{ {d}^{2} + {d}^{2} {k}^{2} + {b}^{2} + {b}^{2} {k}^{2} }{b^{2}k^{2}d^{2}} \right]\\ \end{gathered}$

$\begin{gathered}\\\tt \: \: \red{ \longrightarrow \: \:{d}^{2} + {d}^{2} {k}^{2} + {b}^{2} + {b}^{2} {k}^{2}}\\ \end{gathered}$

$\begin{gathered}\\\tt \: \: \red{ \longrightarrow\: \:{d}^{2} +{(dk)}^{2} + {b}^{2} + {(bk)}^{2}}\\ \end{gathered}$

$\begin{gathered}\\\tt \: \: \red{ \longrightarrow \: \:{d}^{2} +c^{2} + {b}^{2} +a^{2}}\\ \end{gathered}$

$\begin{gathered}\\\tt \: \: \red{ \longrightarrow \: \:R.H.S}\\ \end{gathered}$

✠Hence Proved

Answer 2

$\Huge{\texttt{{{\color{Magenta}{⛄A}}{\red{N}}{\purple{S}}{\pink{W}}{\blue{E}}{\green{R}}{\red{♡}}{\purple{࿐⛄}}{\color{pink}{:}}}}}$

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✠Given:-

$\\\tt \longrightarrow\dfrac{a}{b} = \dfrac{c}{d} = k(const)$

$\\\tt \longrightarrow \: a= bk \: \: and \: \: c = dk$

$\\\tt \: \: \longrightarrow \: \: abcd \left( \dfrac{1}{a^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{c^{2}}+ \dfrac{1}{d^{2}} \right)$

✠Now put the values –

$\\\tt \: \: \longrightarrow \: \: (bk)b(dk)d \left( \dfrac{1}{(bk)^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{(dk)^{2}}+ \dfrac{1}{d^{2}} \right)$

$\\\tt \: \: \longrightarrow \: \: b^{2} {d}^{2} k^{2} \left( \dfrac{1}{(bk)^{2} } + \dfrac{1}{b^{2}}+ \dfrac{1}{(dk)^{2}}+ \dfrac{1}{d^{2}} \right)$

$\\\tt \: \: \longrightarrow \: \: b^{2} {d}^{2} k^{2} \left[\dfrac{ {d}^{2} + {d}^{2} {k}^{2} + {b}^{2} + {b}^{2} {k}^{2} }{b^{2}k^{2}d^{2}} \right]$

$\\\tt \: \: \orange{ \longrightarrow \: \:{d}^{2} + {d}^{2} {k}^{2} + {b}^{2} + {b}^{2} {k}^{2}}$

$\\\tt \: \: \orange{ \longrightarrow\: \:{d}^{2} +{(dk)}^{2} + {b}^{2} + {(bk)}^{2}}$

$\\\tt \: \: \orange{ \longrightarrow \: \:{d}^{2} +c^{2} + {b}^{2} +a^{2}}$

$\\\tt \: \: \orange{ \longrightarrow \: \:R.H.S}$

✠Hence Proved

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$\huge\red{\boxed{\orange{\mathcal{{{\fcolorbox{red}{i}{{\red{@Master}}}}}}}}}$