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

- highschoolmom2010

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- phi

how far did you get ?

- highschoolmom2010

no where im lost

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.