Geography, asked by supriyamaity478, 5 months ago

৩.২০ বন্দরের পশ্চাদভূমি কাকে বলে?/​

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Answered by Anonymous
0

Answer:

\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))</p><p>  (x))(sin(x)+cos(x))(sin(x)−cos(x))</p><p>\large =\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{\sin ^2(x)-\cos ^2(x)}

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