Verify that each equation is an identity by showing that each expression on the left simplifies to 1.

- highschoolmom2010

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- phi

how far did you get ?

- highschoolmom2010

no where im lost

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## More answers

- ganeshie8

take left hand side expression,
plugin tanA = sinA/cosA

- highschoolmom2010

im still really confused

- ganeshie8

did they teach you ' tan = sin/cos' yet ?

- highschoolmom2010

nope

- highschoolmom2010

so im learning it here

- ganeshie8

ok, dont learn it here. wait until they start them officially
lets do it the known way

- highschoolmom2010

ok

- phi

they may want you to use the definitions of sin(A)= opp/adj = a/c
and tan(A)= a/b

- ganeshie8

|dw:1376501404557:dw|

- ganeshie8

exactly,
look at above triangle, first find sinA, and tanA

- highschoolmom2010

|dw:1376501580565:dw|

- ganeshie8

yes, find tanA also, and plug them in left side of given expression

- highschoolmom2010

assuming that is alpha....

- ganeshie8

uhh this time, thats 'a', sorry

- phi

they are using lower case for the side and capital for the angle opposite that side

- highschoolmom2010

\[\tan A=\frac{ a }{ b }\]
oh then let me fix sin A

- highschoolmom2010

\[\sin A=\frac{ a }{ c }\]

- phi

now what is
\[ \frac{1}{\sin A} \]

- highschoolmom2010

@phi im not sure

- ganeshie8

plug them directly, we can simplify after pluggin in may be

- highschoolmom2010

\[\frac{ 1 }{ \frac{ a }{ c} }\]

- phi

the short way is "flip" both sides of
sin A = a/c
the long way is go through these steps
multiply both sides by c:
c sin A = a
divide both sides by a
(c sin A)/a = 1
divide both sides by sin A
c/a = 1/ sin A

- phi

or do what you did , and multiply top and bottom by c/a

- highschoolmom2010

that ^^kinda confused me

- phi

\[ \frac{ 1\cdot \frac{c}{a} }{ \frac{ a }{ c} \cdot \frac{c}{a}} \]
the idea is you can multiply top and bottom by the same number (or fraction)
now simplify

- highschoolmom2010

\[\frac{ \frac{ c }{ a } }{ 1 }\]

- phi

yes, and anything divided by 1 is itself

- highschoolmom2010

co the left side =c/a

- phi

yes. (and remember the short cut)
if you have
a/b = c/d
then
b/a = d/c
(flip both sides) in this case
sin A = a/c
1/sinA = c/a

- highschoolmom2010

so*

- highschoolmom2010

right side = b/a then

- phi

now what is
\[ \frac{1}{\sin^2 A} =\frac{1}{\sin A} \cdot \frac{1}{\sin A} \]

- highschoolmom2010

no clue

- phi

you know 1/sin A = c/a
so you know 1/sin A * 1/sinA = (c/a)*(c/a)
\[ \frac{c}{a}\cdot \frac{c}{a}\]

- phi

which is top times top and bottom times bottom

- phi

lost ?

- phi

another way to do this is remember
\[ \sin^2 A \text{ means } \sin A \cdot \sin A \]
you know sin A= a/c so
\[ \sin^2 A \text{ means } \sin A \cdot \sin A = \frac{a}{c}\cdot \frac{a}{c}=\frac{a\cdot a}{c\cdot c}=\frac{a^2}{c^2}\]

- phi

and
\[ \frac{1}{\sin^2 A} = \frac{c^2}{a^2} \]
using the flip rule

- phi

I should mention that
\[ \left(\sin A \right)^2 \text{ is often written as }\sin^2 A \]
then mean the same thing. (sin A) times itself: (sin A)(sin A)

- highschoolmom2010

im following and writing too

- phi

*they mean...

- phi

so far you have
\[ \frac{1}{(\sin A)^2 }= \frac{c^2}{a^2} \]
now what is 1/(tan A)^2 ?

- highschoolmom2010

1 sec ill write it out on paper and then ^^

- highschoolmom2010

\[\frac{ 1 }{ \tan A^2} \]=\[\frac{ b^2 }{ a^2 }\]

- highschoolmom2010

well i wrote tan A ^2 wrong

- phi

yes, so the equation is
\[ \frac{c^2}{a^2}- \frac{b^2}{a^2} = 1 \]
re: tan A^2 yes, you should put parens in (tan A)^2. but we know what you mean
now, the fractions on the left side have a common denominator
can you re-write as one fraction ?

- highschoolmom2010

\[\frac{ c^2-b^2 }{ a }=1\]

- phi

* a^2 in the bottom

- highschoolmom2010

i knew i forgot something

- phi

now, what is the pythagoras equation for this problem ?

- highschoolmom2010

what does that mean

- phi

do you know the pythagorean theorem ?

- highschoolmom2010

oh yea
a^2+b^2=c^2

- phi

and what do you get if you subtract b^2 from both sides ?

- highschoolmom2010

c^2/a^2=1-b^2

- phi

no, that is too complicated. start with
a^2+b^2=c^2
and write -b^2 on both sides and simplify

- highschoolmom2010

a^2+b^2-b^2=c^2-b^2
a^2=c^2-b^2

- phi

now use that in your equation
\[\frac{ c^2-b^2 }{ a^2 }=1 \]
I would replace the top

- phi

in other words c^2 -b^2 is the same as a^2
(by pythagoras)

- highschoolmom2010

oh

- highschoolmom2010

so what do we do now.....

- phi

the idea is that c^2-b^2 = a^2 so replace c^2-b^2 with a^2 in your equation
\[ \frac{ c^2-b^2 }{ a^2 }=1 \]

- highschoolmom2010

\[\frac{ a^2 }{ a^2 }=1\]

- phi

and something divided by itself is 1
i.e. a^2 divided by itself is 1
you get
1=1
and you are done. You showed both sides are equal

- highschoolmom2010

oh
ok

- highschoolmom2010

ty @phi & @ganeshie8

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