MEDALS!!! Simplify the expression: 1/1+cot^2x a.sec^2x b.csc^2x c.sin^2x d.cos^2x e.tan^2x
rewrite cot^2x in terms of sin cos cotx= ??
there i fixed it
yep okay thanks. so how would you write cotx in terms sin cos ?
idk
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 \[\sin x= \frac{1}{cscx}~~~~~~~~\cos x= \frac{1}{secx}~~~~~~~~\tan =\frac{1}{cotx}=\frac{\sin x}{cosx}\]
if tanx = sinx/cosx then cot x= cosx/sinx
that's how you should rewrite ratios in terms of sin ,cos so rewrite cot^2x as cos^2x over sin^2x
\[\large\rm \frac{ 1 }{ \color{ReD}{1+\frac{ \cos^2x }{ \sin^2x } }}\] simplify the red part first (find the common denominator )
so is it c
what's the common denominator ??
idk
so then how did you get C ??
i saw sin^2x as the denominator
\[ \large\rm 1+\frac{ a }{ b }\] 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 \[\frac{ b+a }{ b }\]
do the same thing with \[1+\frac{\cos^2x}{\sin^2x}\]
sin^2x + cos^2x sin^2x
so d
do you mean over sin^2x ?? \[\frac{ 1 }{ \frac{ \sin^2x+\cos^2x }{ \sin^2x } }\]
yes
so what is the famous identity ?? sin^2x +cos^2x = ??
idk how you type it
thats fine but you can use VVVequation tool :)
\[\large\rm \frac{ 1 }{ \frac{ \color{Red}{\sin^2x+\cos^2x} }{ \sin^2x } }\]what is that red part equal to ?
1?
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 !
so sin^2 x
yep
ok thanks
np
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