Question

Value of |PA – PB|, inside an ideal liquid in accelerating container is

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Answer 1

Value of |PA – PB|, inside an ideal liquid in accelerating container

Explanation:

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

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

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

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

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

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

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

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

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

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

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

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}$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))

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

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

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

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

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

$\mathrm{Cancel\:the\:common\:factor:}\:\sin (x)+\cos(x)Cancelthecommonfactor:sin(x)+cos(x)$

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

$\mathrm{Cancel\:the\:common\:factor:}\:\sin (x)-\cos$

$\mathrm{Use\:the\:following\:identity}:\quad \cos ^2(x)+\sin$

$\huge \boxed{\color{red} {\ \huge =1}}$

Answer 2

Answer:

Value of |PA – PB|, inside an ideal liquid in accelerating container

Explanation:

\color{red} {{{\Large {\bf{To\:\:Simplify\::\frac{\sin ^4(x)-\cos ^4(x)}{\sin ^2(x)-\cos ^2(x)}}}}}}ToSimplify:sin2(x)−cos2(x)sin4(x)−cos4(x)

\color{green}{{{\large {\bf{Your\:\:Answer\::\frac{\sin ^4(x)-\cos ^4(x)}{\sin ^2(x)-\cos ^2(x)}=1}}}}}YourAnswer:sin2(x)−cos2(x)sin4(x)−cos4(x)=1

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

\color{blue} {\large {\bf{Factor\:\sin ^4(x)-\cos ^4(x)}}}Factorsin4(x)−cos4(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}Rewritesin4(x)−cos4(x)as(sin2(x))2−(cos2(x))2=(sin2(x))2−(cos2(x))2

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

\color{fuchsia} {\normalsize \sin ^4(x)=(\sin ^2(x))^2}sin4(x)=(sin2(x))2

\color{fuchsia} {\normalsize =(\sin ^2(x))^2-\cos ^4(x)}=(sin2(x))2−cos4(x) =

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

\color{fuchsia} {\normalsize \cos ^4(x)=(\cos ^2(x))^2}cos4(x)=(cos2(x))2

\color{fuchsia} {\normalsize =(\sin ^2(x))^2-(\cos ^2(x))^2}=(sin2(x))2−(cos2(x))2 =

\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}ApplyDifferenceofTwoSquaresFormula: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))(sin2(x))2−(cos2(x))2=(sin2(x)+cos2(x))(sin2(x)−cos2(x))

=(\sin ^2(x)+\cos ^2(x))(\sin ^2(x)-\cos ^2(x))(sin2(x)+cos2(x))(sin2(x)−cos2(x)) =

\color{blue} {\large {\bf{Factor\:\sin ^2(x)-\cos ^2(x)}}}Factorsin2(x)−cos2(x)

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

\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)-\cos (x))sin2(x)−cos2(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))\ \textless \ br /\ \textgreater \ (x))(sin(x)+cos(x))(sin(x)−cos(x))\ \textless \ br /\ \textgreater \ \large =\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{\sin ^2(x)-\cos ^2(x)}=(sin2(x)+cos2(x))(sin(x)+cos(x))(sin(x)−cos(x)) \textless br/ \textgreater (x))(sin(x)+cos(x))(sin(x)−cos(x)) \textless br/ \textgreater =sin2(x)−cos2(x)(sin2(x)+cos2(x))(sin(x)+cos(x))(sin(x)−cos(x)) =

\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}ApplyDifferenceofTwoSquaresFor