I need this please help

Ответ:

Jasten

11.09.2022

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Step-by-step explanation:

$\huge{\bold☘}\mathfrak\pink{\bold{\underline{{ ℘ɧεŋσɱεŋศɭ}}}}{\bold☘}☘$

℘ɧεŋσɱεŋศɭ

☘

$\red{\bold{\underline{\underline{❥Question᎓}}}}$ ❥Question᎓

Q:-Find the value of

$\sqrt{a - b}$

a−b

$\frac{8 + 3 \sqrt{7} }{8 - 3 \sqrt{7} } - \frac{8 - 3 \sqrt{7} }{8 + 3 \sqrt{7} } = a + b \sqrt{7}$

$\huge\tt\underline\blue{⛶Answer⛶ }$

⛶Answer⛶

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⟹ $\bold{ \frac{8 + 3 \sqrt{7} }{8 - 3 \sqrt{7} } - \frac{8 - 3 \sqrt{7} }{8 + 3 \sqrt{7} } = a + b \sqrt{7}}$ ⟹

⟹ $\bold{ \frac{(8 + 3 \sqrt{7})(8 + 3 \sqrt{7}) - [(8 - 3 \sqrt{7} )(8 - 3 \sqrt{7}) ]}{(8 + 3 \sqrt{7} )(8 - 3 \sqrt{7} )}}$ ⟹

∵ $\bold{ {(a + b)}^{2} - {(a - b)}^{2}$ =

⟹ $\bold{\frac{4 \times 8 \sqrt{7} }{ {(8)}^{2} - {(3 \sqrt{7}) }^{2} } = a + b \sqrt{7} }$ ⟹

⟹ $\bold{\frac{32 \sqrt{7} }{64 - 63} = a + b \sqrt{7}}$ ⟹

64−63

=a+b

⟹ $\bold{ \frac{32 \sqrt{7} }{1} = a + b \sqrt{7}}$ ⟹

{7} = a + b \sqrt{7}}[/tex]⟹32

⟹ $\bold{0 + 32 \sqrt{7} = a + b \sqrt{7} }$ ⟹0+32

$\bold{\red{On \:comparing\: both \:sides:-}}$ Oncomparingbothsides:−

⟹ $\bold{a = 0 \: and \: b = 32}$ ⟹a=0andb=32

⟹ $\bold{now \: \sqrt{a - b} = \sqrt{0 - 32} = \sqrt{ - 32 } = \sqrt{ - 2 \times 16} = 4 \sqrt{ - 2} = 4 \sqrt{2} i \: [( {i}^{2} = - 1)}$ ⟹now

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нσρє ıт нєłρs yσυ

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тнαηkyσυ $\color{red} {{{\Large {\bf{To\:\:Simplify\::\frac{\sin ^4(x)-\cos ^4(x)}{\sin ^2(x)-\cos ^2(x)}}}}}}$

$\color{green}{{{\large {\bf{Your\:\:Answer\::\frac{\sin ^4(x)-\cos ^4(x)}{\sin ^2(x)-\cos ^2(x)}=1}}}}}$

$\color{yellow} {\Huge {\sf{Solution:}}}$

$\color{blue} {\large {\bf{Factor\:\sin ^4(x)-\cos ^4(x)}}}$

$\tt \color{blue} {\mathrm{Rewrite\:}\sin ^4(x)-\cos ^4(x)\mathrm{\:as\:}(\sin ^2(x))^2-(\cos ^2(x))^2=(\sin ^2(x))^2-(\cos ^2(x))^2}$

$\color{fuchsia} {\normalsize {\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^c}}$

$\color{fuchsia} {\normalsize \sin ^4(x)=(\sin ^2(x))^2}$

$\color{fuchsia} {\normalsize =(\sin ^2(x))^2-\cos ^4(x)}$ =

$\color{fuchsia} {\normalsize \mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^c}$

$\color{fuchsia} {\normalsize \cos ^4(x)=(\cos ^2(x))^2}$

$\color{fuchsia} {\normalsize =(\sin ^2(x))^2-(\cos ^2(x))^2}$ =

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}$ x^2-y^2=(x+y)(x-y)

$(\sin ^2(x))^2-(\cos ^2(x))^2=(\sin ^2(x)+\cos ^2(x))(\sin ^2(x)-\cos ^2(x))$

= $(\sin ^2(x)+\cos ^2(x))(\sin ^2(x)-\cos ^2(x))$ =

$\color{blue} {\large {\bf{Factor\:\sin ^2(x)-\cos ^2(x)}}}$

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}$ x^2-y^2=(x+y)(x-y)

$\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)-\cos (x))$

(x)=(sin(x)+cos(x))(sin(x)−cos(x))

=(\sin (x)+\cos (x))(\sin (x)-\cos (x))=(sin(x)+cos(x))(sin(x)−cos(x))

$\large=(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x)) (x))(sin(x)+cos(x))(sin(x)−cos(x))\large =\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{\sin ^2(x)-\cos ^2(x)}$ =

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}$ x^2-y^2=(x+y)(x-y)

$\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)$

$(x)=(sin(x)+cos(x))(sin(x)−cos(x))=\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}$ =

$\mathrm{Cancel\:}\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}:\quad \sin ^2(x)+\cos ^2(x)Cancel (sin(x)+cos(x))(sin(x)−cos(x))$