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Mathematics 19 Online
OpenStudy (xenesis98):

MEDALS!!! Simplify the expression: 1/1+cot^2x a.sec^2x b.csc^2x c.sin^2x d.cos^2x e.tan^2x

Nnesha (nnesha):

rewrite cot^2x in terms of sin cos cotx= ??

OpenStudy (xenesis98):

there i fixed it

Nnesha (nnesha):

yep okay thanks. so how would you write cotx in terms sin cos ?

OpenStudy (xenesis98):

idk

Nnesha (nnesha):

here is good website where you can find all trig ratios,identities, formulas scroll down where it says `Relations Between Trigonometric Functions` sin is reciprocal of 1/csc \[\sin x= \frac{1}{cscx}~~~~~~~~\cos x= \frac{1}{secx}~~~~~~~~\tan =\frac{1}{cotx}=\frac{\sin x}{cosx}\]

Nnesha (nnesha):

if tanx = sinx/cosx then cot x= cosx/sinx

Nnesha (nnesha):

that's how you should rewrite ratios in terms of sin ,cos so rewrite cot^2x as cos^2x over sin^2x

Nnesha (nnesha):

\[\large\rm \frac{ 1 }{ \color{ReD}{1+\frac{ \cos^2x }{ \sin^2x } }}\] simplify the red part first (find the common denominator )

OpenStudy (xenesis98):

so is it c

Nnesha (nnesha):

what's the common denominator ??

OpenStudy (xenesis98):

idk

Nnesha (nnesha):

so then how did you get C ??

OpenStudy (xenesis98):

i saw sin^2x as the denominator

Nnesha (nnesha):

\[ \large\rm 1+\frac{ a }{ b }\] which is same as \[ \large\rm \frac{1}{1}+\frac{ a }{ b }\] to find the common denominator the first fraction by b/b \[\frac{ b }{ b } \cdot \frac{ 1 }{ 1 } +\frac{ a }{ b }=\frac{ b }{ b }+\frac{a}{b}\] now i have the same denominator i can rewrite it as single fraction \[\frac{ b+a }{ b }\]

Nnesha (nnesha):

do the same thing with \[1+\frac{\cos^2x}{\sin^2x}\]

OpenStudy (xenesis98):

sin^2x + cos^2x sin^2x

OpenStudy (xenesis98):

so d

Nnesha (nnesha):

do you mean over sin^2x ?? \[\frac{ 1 }{ \frac{ \sin^2x+\cos^2x }{ \sin^2x } }\]

OpenStudy (xenesis98):

yes

Nnesha (nnesha):

so what is the famous identity ?? sin^2x +cos^2x = ??

OpenStudy (xenesis98):

idk how you type it

Nnesha (nnesha):

thats fine but you can use VVVequation tool :)

Nnesha (nnesha):

\[\large\rm \frac{ 1 }{ \frac{ \color{Red}{\sin^2x+\cos^2x} }{ \sin^2x } }\]what is that red part equal to ?

OpenStudy (xenesis98):

1?

Nnesha (nnesha):

right \[\large\rm \frac{ 1 }{ \frac{ \color{Red}{1} }{ \sin^2x } }\] \[\large\rm \frac{ 1 }{ \frac{ a }{ b } } =1 \cdot \frac{ b }{ a }\] multiply the top with the reciprocal of the bottom fraction that's it !

OpenStudy (xenesis98):

so sin^2 x

Nnesha (nnesha):

yep

OpenStudy (xenesis98):

ok thanks

Nnesha (nnesha):

np

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