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1) \(\frac{1-3cos\theta-4cos^2\theta}{sin^2\theta}=\frac{1-4cos\theta}{1-cos\theta}\)

2) \(tan\theta=\frac{1+tan\theta}{1+cot\theta}\)

AdamTaurus  Apr 1, 2018
edited by AdamTaurus  Apr 1, 2018
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 #1
avatar+85673 
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1)

 

1- 3cosA - 4cos^2A                 1  - 4cosA

________________      =     ___________

       sin^2A                               1  - cos A

 

Note that we can factor the numerator on the left as

 

(1 - 4cos A) ( 1 + cos A)

___________________

  1  - cos^2A

 

(1 - 4cos A) (1 + cos A )

____________________

(1 - cos A) ( 1 + cos A)

 

1 - 4cosA

_________       which equals the right side

1  -  cos A

 

 

 

2)

 

tan A   =    1 + tan A

                 ________

                 1 +  cot A

 

One trick to try here, AT, is to write the right side in terms of sines and cosines

So we have

 

 

1 +  sin A /  cos A

_______________       get common denominators on top/bottom

1  + cos A / sin A

 

 

[cos A + sin A} / cos A

___________________      invert the bottom  fraction and multiply

[ sin A + cos A]  /  sin A

 

 

[ sin A + cosA ]           sin A

_____________   *    ____________

     cos A                   [  sin A + cos A]

 

 

sin A       *    [ sin A + cos A]

____             ____________

cos A          [ sin A + cos A ]

 

 

sin A       *  1

____

cos A

 

tan A  *  1   =   tan A     which is the left side     

 

              

 

cool cool cool

CPhill  Apr 1, 2018

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