Question

Verify the identity. Show your work.
1 + sec^2xsin^2x = sec^2x
my work:
os^2x+sin^2x=1
so (if you divide both sides by cos^2x)...
1+tan^2x=sec^2x tan^2x=sec^2x-1

sec^2x/sec^2x-1=csc^2x
sec^2x/tan^2x=csc^2x
(1/cos^2x)/(sin^2x/cos^2x)=csc^2x
(1/cos^2x)(cos^2x/sin^2x)=csc^2x
(1/sin^2x)=csc^2x
csc^2x=csc^2x
it has some error what i dont know
@mathmath333

Answer 1

\(\large\tt \begin{align} \color{black}{
\color{red}{ 1 + sec^2xsin^2x = sec^2x} \\~\\
1 + sec^2xsin^2x \\~\\
=1 + \dfrac{1}{cos^2x}sin^2x \\~\\
=\dfrac{cos^2x + sin^2x}{cos^2x} \\~\\
=\dfrac{1}{cos^2x} \\~\\
=sec^2x
}\end{align}\)

Answer 2

3) Find the vertical asymptotes 2, if any, of the graph of the rational function. Show your work.


f(x) = (x-4)/(x(x-4))

Vertical asymptotes:
x-4
F(x) = ______
X(x-4)

x-4
F(x) = ______  taking common out
X(x-4)


Leaves it f(x)=x
So the answer would be Vertical asymptotes = x=0, or x=4
find the VA's by setting the denominator to 0.
When we get a factor that cancels out, it is not a vertical asymptote.
What is it? Graph it to see what it looks like.

Answer 3

\[1 + \sec^2x\sin^2x = \sec^2x \]\[1+\frac{\sin^2(x)}{\cos^2(x)}=\sec^2(x)\]\[1+\tan^2(x)=\sec^2(x)\]\[\sec^2(x)=\sec^2(x)~\checkmark\]

Answer 4

Known identities:\[\sin^2(x) +\cos^2(x) = 1\]\[1+\tan^2(x) = \sec^2(x)\]\[1+\cot^2(x) = \csc^2(x)\]

Answer 5

For 3.

Vertical asymptotes happen when denominator = 0

Answer 6

\[1 + \sec^2 x (1 - \cos^2 x) = 1 + \sec^2 x - 1 = \sec^2 x\]So many ways...

Answer 7

Mmhmm, trig is wonderful.

Answer 8

thanx dear @mathmath333 already help me in that one can u help me solve other Q's?

Answer 9

@Jhannybean i am following ur explanation on Q-3

Answer 10

\[x(x-4)=0\\\text{solve forvertical asymptote}\]

Answer 11

that is the denominator part

Answer 12

Sorry, distracted.

Answer 13

there should be 2 asymptote

Answer 14

yes i solve it and got x=0 and x=4

Answer 15

Yeah, that's what I got as well.

Answer 16

but my teacher said : You find the VA's by setting the denominator to 0.
When we get a factor that cancels out, it is not a vertical asymptote.
What is it? Graph it to see what it looks like. what i did wrong?

Answer 17

@jhannybean
@mathmath333

Answer 18

@Nnesha could u help me in above question

Answer 19

a factor that cancels out is what you call a removable discontinuity,

Answer 20

so what step is missing plzz help