Question

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Answer 1

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:a : b \: = \: c : d$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:( {a}^{2} - {c}^{2})( {b}^{2} - {d}^{2}) = {(ab - cd)}^{2} \quad$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:a : b \: = \: c : d$

It means,

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{c}{d}$

Let assume that

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{c}{d} = k$

So,

$\rm :\implies\:a = bk$

and

$\rm :\implies\:c = dk$

Now,

Consider,

$\rm :\longmapsto\:( {a}^{2} - {c}^{2})( {b}^{2} - {d}^{2})$

On substituting the values of a and c, we get

$\rm \: = \: \: \:( {(bk)}^{2} - {(dk)}^{2})( {b}^{2} - {d}^{2})$

$\rm \: = \: \: \:( {b}^{2} {k}^{2} - {d}^{2} {k}^{2})( {b}^{2} - {d}^{2})$

$\rm \: = \: \: \: {k}^{2} ( {b}^{2} - {d}^{2})( {b}^{2} - {d}^{2})$

$\rm \: = \: \: \: {k}^{2} ( {b}^{2} - {d}^{2})^{2}$

can be rewritten as

$\rm \: = \: \: \: {k}^{2} ( {b}.b - {d}.d)^{2}$

$\rm \: = \: \: \: {k}^{2} \bigg( {b}.\dfrac{a}{k} - {d}.\dfrac{c}{k} \bigg)^{2}$

$\red{\bigg \{ \because \: \:\dfrac{a}{b} = \dfrac{c}{d} = k\rm \implies\:b = \dfrac{a}{k} \: and \: d = \dfrac{c}{k} \bigg\}}\:$

$\rm \: = \: \: \: {k}^{2} \bigg(\dfrac{ab - cd}{k} \bigg)^{2}$

$\rm \: = \: \: {k}^{2} \times \dfrac{ {(ab - cd)}^{2} }{ {k}^{2} }$

$\rm \: = \: \: (ab - cd)^{2}$

Hence,

$\boxed{ \bf :\longmapsto\:( {a}^{2} - {c}^{2})( {b}^{2} - {d}^{2}) = {(ab - cd)}^{2} \quad}$

Additional Information :-

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{c}{d}, \: then \:$

$\rm :\longmapsto\:\dfrac{a}{c} = \dfrac{b}{d} \: is \: called \: alternendo$

$\rm :\longmapsto\:\dfrac{b}{a} = \dfrac{d}{c} \: is \: called \: invertendo$

$\rm :\longmapsto\:\dfrac{a + b}{b} = \dfrac{c + d}{d} \: is \: called \: componendo$

$\rm :\longmapsto\:\dfrac{a - b}{b} = \dfrac{c - d}{d} \: is \: called \: dividendo$