Question

(1+cosx)(1+secx)=sinxtanx
prove the following trig identity.

Answer 1

Hello @rl88

Answer 2

Hey!

Answer 3

Help pleasee! :D

Answer 4

Did you try to do that first?

Answer 5

\[(1+a)(1+\frac{1}{a})\] is the first step

Answer 6

or I will say that distributive property is the first hint.

Answer 7

@rl88 do you know distributive property?

Answer 8

what is distributive property?

Answer 9

(a+b)(c+d) = ?

Answer 10

actually, now that i look more carefully, i am not sure it is true

Answer 11

@satellite73 will help you @rl88 I am not feeling well. Also welcome to openstudy @rl88

Answer 12

thanks!

Answer 13

oh maybe it is true, hmm

Answer 14

i'm confused...

Answer 15

i think this is false is what i am saying

Answer 16

suppose \(x=0\)
then the left hand side is \(1+\cos(0))(1+\sec(0))=(1+1)(1+1)=4\) but the right hand side is \(\sin(0)\tan(0)=0\times 0=0\) so this is not an identity

Answer 17

look carefully at the problem and make sure you copied it correctly, because this one is not true

Answer 18

hmm no, its right, theres another question for example 1-2sin^2x=2cos^2x-1 the rearrangement of that question gets you a true statement but that statement alone doesn't if you substitute x for 0.

Answer 19

an "identity" means true for all values of \(x\)
like \(2(x-4)=2x-8\) is an identity
it cannot only be true sometimes

Answer 20

ohh

Answer 21

if you graph both of them above, you will see that they are not the same, but i any case you do not need to do that, because letting \(x=0\) and getting \(4=0\) proves that it is not an identity

Answer 22

hmm...OK thanks!

Answer 23

i know what it is supposed to be i think

Answer 24

\((1-\cos(x))(1+\sec(x))\) might be right

Answer 25

maybe that was the actual problem, show
\[(1-\cos(x))(1+\sec(x))=\sin(x)\tan(x)\] that i think we can do

Answer 26

might that have been the question?

Answer 27

maybe it was a typo on my page? it could have been? how would that be done?

Answer 28

it is easier to write the algebra if you replace \(\cos(x)\) by \(a\)
you get
\[(1-a)(1+\frac{1}{a})=1+\frac{1}{a}-a-1=\frac{1}{a}-a=\frac{1-a^2}{a}\]

Answer 29

then since \(a=\cos(x)\) the numerator is \(1-\cos^2(x)=\sin^2(x)\) making the whole fraction
\[\frac{\sin^2(x)}{\cos(x)}=\sin(x)\tan(x)\]

Answer 30

Hey is anyone here good in English and knows a lot about the book Hamlet please hmu with a message! =) ty

Answer 31

ohh...i dont get the last bit? how does sin^2x/cosx =sinxtanx?