anonymous
  • anonymous
prove the following identities: cot(A-B)=(cot A cot B+1)/(cot B-cot A)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
CGGURUMANJUNATH
  • CGGURUMANJUNATH
|dw:1377596937013:dw|
CGGURUMANJUNATH
  • CGGURUMANJUNATH
is it clear ?
CGGURUMANJUNATH
  • CGGURUMANJUNATH
@chengyen r u there ?

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CGGURUMANJUNATH
  • CGGURUMANJUNATH
\[\cot \left( A-B \right)=\cos \left( A-B \right)/\sin \left( A-B \right)\]
CGGURUMANJUNATH
  • CGGURUMANJUNATH
NOW EXPAND COS(A-B) AND SIN(A-B) USING TRIGONOMETRIC FORMULAE AND THEN DIVIDE NUMERATOR OVER DENOMINATOR BY PROODUCT OF SIN A AND SIN B TO GET THE FINAL ANSWER.
anonymous
  • anonymous
|dw:1377597630518:dw|
anonymous
  • anonymous
|dw:1377597795095:dw|
anonymous
  • anonymous
then i dun know alrdy~
CGGURUMANJUNATH
  • CGGURUMANJUNATH
very good
CGGURUMANJUNATH
  • CGGURUMANJUNATH
lhs step is correct
CGGURUMANJUNATH
  • CGGURUMANJUNATH
now divide each term of NUMERATOR OVER DENOMINATOR BY PROODUCT OF SIN A AND SIN B
CGGURUMANJUNATH
  • CGGURUMANJUNATH
you cannot cancel cos B LIKE THAT
CGGURUMANJUNATH
  • CGGURUMANJUNATH
|dw:1377598245671:dw|
CGGURUMANJUNATH
  • CGGURUMANJUNATH
ARE YOU CLEAR WITH THE NUMERATOR ?
CGGURUMANJUNATH
  • CGGURUMANJUNATH
\[(\cos A \cos B+\sin A \sin B)/(\sin A \cos B-\cos A \sin B)\]
CGGURUMANJUNATH
  • CGGURUMANJUNATH
\[(cosAcosB/sinAsinB+sinAsinB/sinAsinB)/(sinAcosB/sinAsinB−cosAsinB/sinAsinB)\]
CGGURUMANJUNATH
  • CGGURUMANJUNATH
IS IT CLEAR ?
CGGURUMANJUNATH
  • CGGURUMANJUNATH
HELLO ! chengyen
anonymous
  • anonymous
wait,wait ,let me see first..
CGGURUMANJUNATH
  • CGGURUMANJUNATH
OK
CGGURUMANJUNATH
  • CGGURUMANJUNATH
THEN USE COS/SIN AS COT AND CANCEL THE LIKE TERMS.SIMPLE IS'NT IT ?
anonymous
  • anonymous
|dw:1377599111735:dw| ?? is like this?
CGGURUMANJUNATH
  • CGGURUMANJUNATH
|dw:1377599323639:dw|
CGGURUMANJUNATH
  • CGGURUMANJUNATH
READ THE 16TH CONVERSATION.
CGGURUMANJUNATH
  • CGGURUMANJUNATH
|dw:1377599711923:dw|
anonymous
  • anonymous
sorry,i see wrong..
CGGURUMANJUNATH
  • CGGURUMANJUNATH
|dw:1377599786166:dw|
CGGURUMANJUNATH
  • CGGURUMANJUNATH
|dw:1377599901592:dw|
CGGURUMANJUNATH
  • CGGURUMANJUNATH
yes what step you had written above was wrong but the above 2 steps what i wrote are correct.DID U GET IT NOW ?
anonymous
  • anonymous
calm down,dun angry because will fast old,wakaka,whatever thank you^^yes,i get!
CGGURUMANJUNATH
  • CGGURUMANJUNATH
will fast old,wakaka ? ? ?
anonymous
  • anonymous
you dun know??
CGGURUMANJUNATH
  • CGGURUMANJUNATH
WHY SHOULD I BE ANGRY ?
CGGURUMANJUNATH
  • CGGURUMANJUNATH
WHAT ? you dun know??
CGGURUMANJUNATH
  • CGGURUMANJUNATH
I AM JUST INTERESTED TO HELP YOU.
anonymous
  • anonymous
because i an stupid one.oh,then i misunderstood it.
CGGURUMANJUNATH
  • CGGURUMANJUNATH
WHO TOLD UR ARE STUPID YOU ARE A INTELLIGENT LEARNER.
anonymous
  • anonymous
you can continue teach other,i will remember'I AM A INTELLIGENT LEARNER'.wah,so meaningful,thank you. :)
CGGURUMANJUNATH
  • CGGURUMANJUNATH
you're welcome.
DebbieG
  • DebbieG
@chengyen , do you understand why you absolutely cannot cancel like this? --> |dw:1377600815909:dw| I ask because I see this a LOT from students, and it's something that is important to understand. Look at it this way: IF you could do this, then wouldn't it also be true that: \[\Large \frac{ 15 }{5 }=\frac{ 12+\cancel{3}}{ 2+\cancel3 }=\frac{ 12}{ 2 }+1=6+1=7\] Soooo...... \(\Large \dfrac{ 15 }{5 }=7\) right? :) Obviously this is wrong, because that factoring is wrong, because the 3's are not COMMON FACTORS in the num'r and den'r, they are common factors only to the SECOND TERM of the num'r and den'r, but that doesn't allow us to cancel them! My example is just plain silly of course..... but it is exactly the same kind of canceling that you did above with the sin(a)sin(b).
anonymous
  • anonymous
emm......oh,i know how alrdy,thanks^^
DebbieG
  • DebbieG
OK, just wanted to check since you cancelled incorrectly, in similar ways, twice. :)

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