Question

proving identities:
(1-3cos-4cos^2)/(sin^2) = (1-4cos)/(1-cos)

Answer 1

@UnkleRhaukus

Answer 2

let U=cos x
1-3cos-4cos^2= 1-3U-4U^2=(1-4U)(1+U)
___> 1-3cos-4cos^2=(1-4cos)(1+cos)
sin^2 x+cos^2 x=1 _____> sin^2 x=1-cos^2 x
(1-3cos-4cos^2)/(sin^2)=(1-4cos)(1+cos)/(1-cos^2 x)
=(1-4cos)(1+cos)/(1-cos x)(1+cos x)
=(1-4cos)/(1-cos)

Answer 3

i dont understand @mustafa2014

Answer 4

\[\frac{ 1 - 3\cos \theta - 4\cos^{2}\theta }{ \sin^{2}\theta } = \frac{ 1 - 4\cos \theta }{ 1 - \cos \theta }\]

before we solve this... let's just remember this
\[\cos^{2} \theta + \sin^{2} \theta =1\]
which means
\[\sin^{2}\theta = 1 - \cos^{2}\theta = ( 1- \cos \theta) ( 1 + \cos \theta) - difference \ of \ 2 \ squares \]
now lets consider the left hand side....
\[\frac{ 1 - 3\cos \theta - 4\cos^{2} \theta }{ \sin^{2} \theta }\]
factorize the upper part
\[\frac{ (1- 4\cos \theta)( 1 + \cos \theta) }{ \sin^{2} \theta }\]
now expand sin^2 theta using the expression i mention above
\[\frac{ (1 - 4\cos \theta)(1 + \cos \theta) }{ (1-\cos \theta)(1 + \cos \theta) }\]

now u have the answer.... ancel out the ( 1+ cos theta) part from both numerator ande denominator

and u will have
\[\frac{ (1 - 4\cos \theta) }{( 1 - \cos \theta) }\]

hope this will help ya!! and srry for taking a long time :)