if 15 cotA =7 then find the value of cosec A
Answers
Answer:
QUESTION✏ :-
if tanA=15/7,find the value of cosecA+cotA
ANSWER✏✏✏:-
\color{red} {{{\Large {\bf{To\:\:Simplify\::\frac{\sin ^4(x)-\cos ^4(x)}{\sin ^2(x)-\cos ^2(x)}}}}}}ToSimplify:
sin
2
(x)−cos
2
(x)
sin
4
(x)−cos
4
(x)
\color{green}{{{\large {\bf{Your\:\:Answer\::\frac{\sin ^4(x)-\cos ^4(x)}{\sin ^2(x)-\cos ^2(x)}=1}}}}}YourAnswer:
sin
2
(x)−cos
2
(x)
sin
4
(x)−cos
4
(x)
=1
\color{yellow} {\Huge {\sf{Solution:}}}Solution:
\color{blue} {\large {\bf{Factor\:\sin ^4(x)-\cos ^4(x)}}}Factorsin
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}Rewritesin
4
(x)−cos
4
(x)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}}Applyexponentrule:a
bc
=(a
b
)
c
\color{fuchsia} {\normalsize \sin ^4(x)=(\sin ^2(x))^2}sin
4
(x)=(sin
2
(x))
2
\color{fuchsia} {\normalsize =(\sin ^2(x))^2-\cos ^4(x)}=(sin
2
(x))
2
−cos
4
(x) =
\color{fuchsia} {\normalsize \mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^c}Applyexponentrule:a
bc
=(a
b
)
c
\color{fuchsia} {\normalsize \cos ^4(x)=(\cos ^2(x))^2}cos
4
(x)=(cos
2
(x))
2
\color{fuchsia} {\normalsize =(\sin ^2(x))^2-(\cos ^2(x))^2}=(sin
2
(x))
2
−(cos
2
(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))(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))(sin
2
(x)+cos
2
(x))(sin
2
(x)−cos
2
(x)) =
\color{blue} {\large {\bf{Factor\:\sin ^2(x)-\cos ^2(x)}}}Factorsin
2
(x)−cos
2
(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))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)}=(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 =
sin
2
(x)−cos
2
(x)
(sin
2
(x)+cos
2
(x))(sin(x)+cos(x))(sin(x)−cos(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)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))}(x)=(sin(x)+cos(x))(sin(x)−cos(x)) \textless br/ \textgreater =
(sin(x)+cos(x))(sin(x)−cos(x))
(sin
2
(x)+cos
2
(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))Cancel
(sin(x)+cos(x))(sin(x)−cos(x))
(sin
2
(x)+cos
2
(x))(sin(x)+cos(x))(sin(x)−cos(x))
: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)Cancelthecommonfactor:sin(x)+cos(x)Cancelthecommonfactor:sin(x)+cos(x)
=\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)-\cos (x))}{\sin (x)-\cos (x)}
sin(x)−cos(x)
(sin
2
(x)+cos
2
(x))(sin(x)−cos(x))
=
\mathrm{Cancel\:the\:common\:factor:}\:\sin (x)-\cosCancelthecommonfactor:sin(x)−cos
\mathrm{Use\:the\:following\:identity}:\quad \cos ^2(x)+\sinUsethefollowingidentity:cos
2
(x)+sin
\huge \boxed{\color{red} {\ \huge =1}}
=1
\huge{Hope\;it\;helps\;u}Hopeithelpsu
Hope it helps ☺️
Please mark it brainliest
Have a good day
