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# Verify Each Identity

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

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

Apr 1, 2018
edited by AdamTaurus  Apr 1, 2018

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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

Apr 1, 2018