Question

Verify, then find which is NOT equivalent : (Picture below)

Answer 1

\[\frac{ cosx }{ sinx }\]

Answer 2

Can you show me how to figure that out?

Answer 3

sure

Answer 4

sorry

Answer 5

\[\sin(\pi/2-x)\csc\]

Answer 6

is this the question

Answer 7

because in the picture the question does not seem right

Answer 8

if this is the question

Answer 9

\[\sin(\frac{ \pi }{2 }-x)=\sin(\frac{ \pi }{ 2 })cosx-\cos(\frac{ \pi }{ 2 })sinx=1.cosx-0.sinx=cosx\]

Answer 10

having this in mind, let us insert it in the original expression

Answer 11

\[\sin(\frac{ \pi }{2 }-x)cscx=cosxcscx=cosx.\frac{ 1 }{ sinx }\]

Answer 12

cosx/sinx=cotx=1/tanx

Answer 13

done

Answer 14

Thank you!
I have another one that maybe you could help me with?

Answer 15

for sure

Answer 16

okay, hold on let me get it..

Answer 17

If you could do it all in one message that would be great so I can follow along easier.

Answer 18

\[\frac{ cscA-sinA }{ \cot ^{2}A }=\frac{\frac{ 1 }{sinA }-sinA }{ \frac{ \cos ^{2}A }{ \sin ^{2}A } }=\frac{ \frac{ 1-\sin ^{2}A }{sinA } }{ \frac{ \cos ^{2}A }{\sin ^{2} A} }\]
\[\frac{ 1-\sin ^{2}A }{sinA }.\frac{ \sin ^{2} A}{ \cos ^{2}A }=\frac{ \cos ^{2}A sinA}{ cosA }=sinAcosA=\frac{ 1 }{2 }\sin(2A)\]

Answer 19

Here is the final answer

Answer 20

So what's the final answer?

Answer 21

\[\frac{ 1 }{ 2 }\sin(2A)\]

Answer 22

does it make sense

Answer 23

uhmm.. yeah sort of. im terrible at this whole simplifying thing. up for a few more?

Answer 24

If you practice, it is not that hard my friend

Answer 25

i got all of the verifying ones but this one..

Answer 26

I am going for launch. If you stay for 30 minutes i will get back to you.

Answer 27

I'll be here all evening. Thank you! Have a good time :)

Answer 28

\[\frac{ \cos \alpha-\cos \beta }{ \sin \alpha+\sin \beta }+\frac{ \sin \alpha-\sin \beta }{ \cos \alpha+\cos \beta}=\]
\[\frac{ \cos ^{2}\alpha-\cos ^{2}\beta+\sin ^{2}\alpha-\sin^2\beta }{(\sin \alpha+\sin \beta)(\cos \alpha +\cos \beta)}\]
=\[\frac{ \cos^2 \alpha+\sin^2\alpha-(\sin^2\beta +\cos^2\beta) }{(\sin \alpha+\sin \beta)(\cos \alpha+\cos \beta) }=\frac{ 1-1 }{(\sin \alpha+\sin \beta) (\cos \alpha+\cos \beta) }=0\]

Answer 29

it is verified

Answer 30

how is the denominator verified? nothing was done to it?

Answer 31

since it does not have any significance to simplify the denominator, we can leave it as it is.

Answer 32

How do you know if it has significance to simplify it or not?

Answer 33

In this particular problem, the purpose is to simplify the right hand side to zero. That tells us the numerator of the RHS has to be siplified to zero. our focus should be mostly on working out the numerator.

Answer 34

ohh okay. thanks.

Answer 35

it is my pleasure