+895 1) \(\frac{1-3cos\theta-4cos^2\theta}{sin^2\theta}=\frac{1-4cos\theta}{1-cos\theta}\)
2) \(tan\theta=\frac{1+tan\theta}{1+cot\theta}\)
+128475 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
