Question

Ver

Answer 1

@pooja195 i really need help

Answer 2

\[\frac{ \tan(x)+\pi }{ 2 }=-\cot(x)\]

Answer 3

please write out how to verify this one please

Answer 4

thinking about it.

Answer 5

i think the question is \[\tan(x) + \frac{\pi}{2} = -\cot(x)\] am i right ?

Answer 6

ah yea, that would make sense. then you could use cofunction identities I believe

Answer 7

or \[\tan(x+ \frac{\pi}{2})=-\cot(x)\]

Answer 8

yes thats correct @Nnesha

Answer 9

yup yup, trying to figure it out now

Answer 10

i hope you know that tanx = sinx/cosx
\[\tan(x+\frac{\pi}{2})= \frac{\sin(x+\frac{\pi}{2})}{\cos(x+\frac{\pi}{2})}\]now you can apply sum/difference formula
which is \[\large\rm \sin(x+y)= sin(x) cos(y)+\cos( x) \sin( y)\]\[\rm \cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)\]

Answer 11

That should work.

Answer 12

@Nnesha got it?

Answer 13

so @Nnesha proved it?

Answer 14

yes.apply sum formula ive posted above

Answer 15

thank you so much. @Nnesha

Answer 16

yw. let me know what you get. i can check your work

Answer 17

if you have any question..feel free to ask

Answer 18

I don't know how to apply it when I try... :-(

Answer 19

@Loser66

Answer 20

Can you post the original question by scanning?

Answer 21

Question: verify the identity: tan(x+π2)=−cot(x)

Answer 22

ok

Answer 23

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha
or \[\tan(x+ \frac{\pi}{2})=-\cot(x)\]
\(\color{blue}{\text{End of Quote}}\)

Answer 24

oops I meant: tan(x+π/2)=−cot(x)

Answer 25

\(tan(x +\pi/2)=\dfrac{sin(x+\pi/2}{cos(x+\pi/2)}\)
OK?

Answer 26

ok

Answer 27

Now, numerator: \(sin(x+pi/2)= sin xcos(\pi/2)+sin(\pi/2) cos x\)
OK?

Answer 28

oh ok! got it so far.

Answer 29

but cos (pi/2) =0 and sin(pi/2)=1, so
numerator = cos x
OK?

Answer 30

ok

Answer 31

Now, denominator \(cos(x+\pi/2) = cos x cos(\pi/2) - sinxsin(\pi/2)\)
again , \(cos(\pi/2) =0, sin(\pi/2) =1\)
hence denominator = -sin x, ok?

Answer 32

ok

Answer 33

numerator/denominator = cos x/-sin x = -cot x
that's it

Answer 34

ah ok. So that proved it all? I see now. Can you check one of my other proofs for me?

Answer 35

ok

Answer 36

thaks! Here it is:

Answer 37

1.) Verify the identity.

quantity one minus sine of x divided by cosine of x equals cosine of x divided by quantity one plus sine of x


-1 - sin x / cos x = cos x / 1 + sin x , then 1 - sin x / cos x X 1 + sin x / 1 + sin x then, 1 - sin^2 x/cos x (1 + sin x) , cos^2 x / cos x (1 + sin x), and then cos x / 1 + sin x.

Answer 38

So I listed the question and then the answer

Answer 39

math notation, please

Answer 40

one minus sine of x,
but you wrote -1-sin(x)
???

Answer 41

the question was: Verify the identity: 1-sinx/cos x = cos x/ 1+sin x.

I wrote: 1 - sin x / cos x = cos x / 1 + sin x , then 1 - sin x / cos x X 1 + sin x / 1 + sin x then, 1 - sin^2 x/cos x (1 + sin x) , cos^2 x / cos x (1 + sin x), and then cos x / 1 + sin x.

Answer 42

this is the trick!!
You have to go from the left hand side to the right hand side, right?
the denominator of the right hand side is (1+sinx)
how to get it from the left hand side?
just multiple both numerator and denominator of the left hand side by it!!