Question

tan2x+sec2x=2

Answer 1

Is it, \(\large\color{black}{ \tan^2x+\sec^2x=2}\) ?

Answer 2

yes

Answer 3

sorry no

Answer 4

it is as posted

Answer 5

then is it, \(\large\color{black}{ \tan(2x)+\sec(2x)=2}\) right?

Answer 6

yes

Answer 7

Connected snapped, sorry

Answer 8

it is not showing that I am typing but I am here

Answer 9

ok

Answer 10

\(\large\color{black}{ \tan(2x)=\frac{\LARGE \sin(2x)}{\LARGE\cos(2x)}=\frac{\LARGE 2\sin(x)\cos(x)}{\LARGE\cos^2(x)-\sin^2(x)}}\)


\(\large\color{black}{ \sec(2x)=\frac{\LARGE 1}{\LARGE\cos(2x)}=\frac{\LARGE1}{\LARGE\cos^2(x)-\sin^2(x)}}\)

Answer 11

you had: \(\large\color{black}{ \tan(2x)+\sec(2x)=2}\) can you re-write your expression.
(you can draw too, if you want).

Answer 12

need help re-writing it?

Answer 13

yes I need help working the problem

Answer 14

i have the answer but cannot arrive at it on my own

Answer 15

okay, so we had: \(\large\color{black}{ \tan(2x)+\sec^2(x)=2}\)


and now get: \(\large\color{black}{ \frac{\LARGE 2\sin(x)\cos(x)}{\LARGE\cos^2(x)-\sin^2(x)}+\frac{\LARGE1}{\LARGE\cos^2(x)-\sin^2(x)}=2}\)

Answer 16

see how I amdoing it?

Answer 17

ok

Answer 18

can you add the fractions?

Answer 19

yes

Answer 20

show me what you get, please

Answer 21

2sinxcosx+1/cos^2x-sin^2x=2

Answer 22

yes;) \(\large\color{black}{ \frac{\LARGE 2\sin(x)\cos(x)+1}{\LARGE\cos^2(x)-\sin^2(x)}=2}\)

Answer 23

\(\large\color{black}{ \frac{\LARGE 2\sin(x)\cos(x)+1}{\LARGE\cos^2(x)-\sin^2(x)}=2}\)

\(\large\color{black}{ 2\sin(x)\cos(x)=2 \times [\cos^2(x)-\sin^2(x)]}\)

Answer 24

can you re-write the RIGHT side in terms of sines only, OR in terms fo cosines only?

Answer 25

\[(~using,~~~~\sin^2x+\cos^2x=1~)\]

Answer 26

2 x (sin^2x + 1)

Answer 27

can you try again please?

Answer 28

-sin^2x+1

Answer 29

\(\large\color{red}{ \cos^2(x)+\sin^2(x)=1}\)
\(\large\color{red}{ \sin^2(x)=1-\cos^2(x)}\)

\(\large\color{black}{ 2\sin(x)\cos(x)=2 \times [\cos^2(x)-\left\{ 1-\cos^2(x) \right\}~]}\)

Answer 30

\(\large\color{black}{ 2\sin(x)\cos(x)=2 \times [\cos^2(x)- 1+\cos^2(x) ]}\)

\(\large\color{black}{ 2\sin(x)\cos(x)=2 \times [2\cos^2(x)- 1]}\)

Answer 31

now, use the rule:
\(\large\color{black}{ \sin(x)=\sqrt{1-\cos^2(x)}}\)

for the left side.

Answer 32

what will your left side be can you tell me?

Answer 33

sin2x

Answer 34

you are combining the left side again, but we need to re-write just in terms of cos(x).

Answer 35

wouldn't you get:

\(\large\color{black}{ 2~\sqrt{1-\cos^2(x)}~~\cos(x)~=2 \times [2\cos^2(x)- 1]}\)

Answer 36

then the 2s cancel:
\(\large\color{black}{ \sqrt{1-\cos^2(x)}~~\cos(x)~=2\cos^2(x)- 1}\)

Answer 37

then substitute something for cos(x). say, let u=cos(x).

Answer 38

what will your equation look after you sub in u?

Answer 39

square root (1-u^2)(u)

Answer 40

yes that is the left side, and the right ?

Answer 41

2u^2 -1

Answer 42

yes

Answer 43

So we get, \(\large\color{black}{ u\sqrt{1-u^2}~=2u^2- 1}\)

Answer 44

we will need to square both sides to get rid of the square root.

Answer 45

\(\large\color{black}{ (~ u\sqrt{1-u^2}~~)^2=(~2u^2- 1~)^2}\)

Answer 46

can you raise the left side to the second power?

Answer 47

\(\large\color{black}{ (~ u\sqrt{1-u^2}~~)^2=(~2u^2- 1~)^2}\)
is same as,

\(\large\color{black}{ (~ u)^2~(\sqrt{1-u^2}~~)^2=(~2u^2- 1~)^2}\)

now can you do the left side?

Answer 48

u^2 (1-u)^2

Answer 49

wait, a square root of a number raised to the second power, = the number (itself) right?

Answer 50

thought so

Answer 51

you got the u^2 right, but \(\large\color{black}{ (\sqrt{1-u^2}~~)^2=1-u^2}\) right?

Answer 52

yea i accidentally closed parenthesi

Answer 53

so what is the left side going to be?

Answer 54

\(\large\color{black}{ (~ u\sqrt{1-u^2}~~)^2=(~2u^2- 1~)^2}\)

\(\large\color{black}{ (~ u)^2~(\sqrt{1-u^2}~~)^2=(~2u^2- 1~)^2}\)

\(\large\color{black}{ u^2(1-u^2)=(~2u^2- 1~)^2}\)

Answer 55

4u+1

Answer 56

that is for what? how do you get to this?

Answer 57

I am not saying it must be wrong, but I just don't understand what you did.

Answer 58

right side...

Answer 59

you are to do this on the right side:

\(\large\color{black}{ u^2(1-u^2)=(~2u^2- 1~)^2}\)

\(\large\color{black}{ u^2(1-u^2)=(2u^2)^2-2(2u^2)(1)+(1)^2}\)


this is a rule: \(\large\color{black}{ (a+b)^2=a^2-2ab+b^2}\) that I am applying.

Answer 60

ok

Answer 61

can you simplify the right side?

Answer 62

\(\large\color{black}{ u^2(1-u^2)=(2u^2)^2- (2)(2u^2)(1)+(1)^2}\)

you can use my code
`\(\large\color{black}{ u^2(1-u^2)=(2u^2)^2- (2)(2u^2)(1)+(1)^2}\)`
(cope paste from the gray)

Answer 63

4u-4u^2+1

Answer 64

the first term is incorrect

Answer 65

\(\large\color{black}{ (2u^2)^2=(2)^2~(u^2)^2=4u^4}\)

Answer 66

\(\large\color{black}{ u^2(1-u^2)=4u^4-4u^2+1}\)

Answer 67

\(\large\color{black}{ u^2-u^4=4u^4-4u^2+1}\)

Answer 68

\(\large\color{black}{ u^2-u^4\color{red}{-u^2+u^4}=4u^4-4u^2+1\color{red}{-u^2+u^4}}\)

\(\large\color{black}{ 0=5u^4-5u^2+1}\)

Answer 69

tell me if you get this so far.

Answer 70

yep

Answer 71

\(\large\color{black}{ 5u^4-5u^2+1=0 }\)

Answer 72

lets substitute for \(\large\color{black}{ u^2 }\).
let \(\large\color{black}{ u^2=w }\)

So we have: \(\large\color{black}{ w=u^2=\cos^2(x) }\)
and so, \(\large\color{black}{ w=\cos^2(x) }\) (when we sub back)

\(\large\color{black}{ 5w^2-5w+1=0 }\)

Answer 73

ok

Answer 74

can you solve this quadratic?

Answer 75

use the quadratic formula.

Answer 76

\(\normalsize\color{ slate }{\Huge{\bbox[5pt, lightcyan ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{-\color{magenta}{b} \pm\sqrt{ \color{magenta}{b} ^2-4 \color{blue}{a} \color{red}{c}}}{2 \color{blue}{a}} }~ }}}\)

when the equation is \(\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\).

Answer 77

\(\large\color{black}{ \color{blue}{5}w^2 \color{magenta}{-5}w+\color{red}{1}=0 }\)

Answer 78

can you do it now?

Answer 79

-(-5w)-squareroot4(5w^2)(1)/2(5w^2)

Answer 80

you are plugging just the coefficients and numbers into the formula for w, not the Ws.