Question

How to verify identities?

Answer 1

Verify the identity.
\[\frac{ cosx }{ 1+sinx }+\frac{ 1+sinx }{ cosx }=2secx\]

Answer 2

The LCD of the left side is cosx(1+sinx)

so I would get each fraction equal to the LCD and then combine the fractions

Answer 3

\[\frac{ \cos^2x }{ \cos(1+sinx) }+\frac{ 1+\sin^2x }{ \cos(1+sinx) }\]

Answer 4

close...


\[\frac{ cosx }{ 1+sinx }+\frac{ 1+sinx }{ cosx }=2secx\]


\[\frac{ cosx }{ cosx }*\frac{ cosx }{ 1+sinx }+\frac{ 1+sinx }{ cosx }=2secx\]


\[\frac{ cos^2x }{ cosx(1+sinx) }+\frac{ 1+sinx }{ cosx }=2secx\]


\[\frac{ cos^2x }{ cosx(1+sinx) }+\frac{ 1+sinx }{ cosx }*\frac{ 1+sinx }{ 1+sinx }=2secx\]


\[\frac{ cos^2x }{ cosx(1+sinx) }+\frac{ (1+sinx)^2 }{ cosx (1+sinx) }=2secx\]


\[\frac{ cos^2x }{ cosx(1+sinx) }+\frac{ 1+2sinx+sin^2x }{ cosx (1+sinx) }=2secx\]

Answer 5

now combine the fractions and simplify as much as possible

Answer 6

Oh okay I did the numerator for the second one wrong. so combining them I would get\[\frac{ 1+2sinx+\sin^2x+cosx }{ cosx(1+sinx) }\]

Answer 7

that cosx should be cos^2x

Answer 8

oh yes whoops

Answer 9

sin^2x + cos^2x = 1 which is an identity, so

\[\frac{ 1+2sinx+sin^2x+cos^x }{ cosx(1+sinx) }\]

turns into

\[\frac{ 1+2sinx+1 }{ cosx(1+sinx) }\]

Answer 10

combine like terms

\[\frac{ 2+2sinx }{ cosx(1+sinx) }\]

Answer 11

what's next?

Answer 12

You would get\[\frac{ 1+sinx }{ \cos }\] I believe

Answer 13

no

Answer 14

\[\frac{ 2+2sinx }{ cosx(1+sinx) }\]

\[\frac{ 2(1+sinx) }{ cosx(1+sinx) }\]


notice that the 1+sinx terms cancel leaving

\[\frac{ 2 }{ cosx }\]

which turns into

\[2secx\]


keep in mind that each step has a = 2secx at the end that never changes

Answer 15

Ohhhh okay I see now.

Answer 16

yeah it's basically a chain of equivalent equations where you start with what you're given and end up with 2secx = 2secx

Answer 17

once you pick a side, you only change that side

Answer 18

Alright so what would I do If it says to verify the identity of something that I already know is an identity? like \[\cot (x-\frac{ \pi }{ 2 })=-tanx\]

Answer 19

No single method works for all identities. However following certain steps might help. To verify an identity, you may start by transforming the more complicated side into the other using basic identities. Or you may transform the two sides into one same expression.

Example 1:
Verify the identity cos x * tan x = sin x

Solution to Example 1:

We start with the left side and transform it into sin x. Use the identity tan x = sin x / cos x in the left side.

cos x * tan x = cos x * (sin x / cos x) = sin x

Example 2:
Verify the identity cot x * sec x * sin x = 1

Solution to Example 2

Use the identities cot x = cos x / sin x and sec x = 1/ cos x in the left side.

cot x * sec x * sin x = (cos x / sin x) * (1/ cos x) * sin x

Simplify to obtain.

(cos x / sin x) * (1/ cos x) * sin x = 1
Example 3:

Verify the identity [ cot x - tan x ] / [sin x * cos] = csc2x - sec2x

Answer 20

use other identities that you know to show that it's true

Answer 21

so use

cotx = 1/tanx

and use

tan(x+y) = (tanx + tany)/(1-tanx*tany)

Answer 22

what would x and y be? 0 and 1

Answer 23

x and pi/2

Answer 24

because of the unit cirlce value of pi/2

Answer 25

but now that I think about it, tan(pi/2) is undefined, so that's a problem

Answer 26

you'll have to do it this way

\[\cot(x - \pi/2) = -\tan(x)\]

\[\frac{\cos(x - \pi/2)}{\sin(x - \pi/2)} = -\tan(x)\]

\[\frac{\cos(x)\cos(\pi/2) + \sin(x)\sin(\pi/2)}{\sin(x)\cos(\pi/2) - \cos(x)\sin(\pi/2)} = -\tan(x)\]

see what's next?

Answer 27

combining?

Answer 28

cos(pi/2) = ???

sin(pi/2) = ???

Answer 29

0 and 1

Answer 30

good

Answer 31

so

\[\frac{\cos(x)\cos(\pi/2) + \sin(x)\sin(\pi/2)}{\sin(x)\cos(\pi/2) - \cos(x)\sin(\pi/2)} = -\tan(x)\]

turns into

\[\frac{\cos(x)*0 + \sin(x)*1}{\sin(x)*0 - \cos(x)*1} = -\tan(x)\]

which becomes

\[\frac{0 + \sin(x)}{0 - \cos(x)} = -\tan(x)\]

\[\frac{\sin(x)}{-\cos(x)} = -\tan(x)\]

\[-\frac{\sin(x)}{\cos(x)} = -\tan(x)\]

\[-\tan(x) = -\tan(x)\]

Answer 32

oh my gosh I see that! thank you so much! I was so confused

Answer 33

you're welcome