How do you prove #sin^2x/ (cos^2 +3cosx+2) = (1-cosx)/ (2+cosx)#?

4 Answers
Apr 16, 2018

As proved.

Explanation:

#sin^2 x / (cos^2x + 3 cos x + 2)#

#=> (1-cos^2 x) / (cos^2 x + cos x + 2 cos x + 2)#

#=> ((1 + cos x) * (1 - cos x)) / (cos x (cos x + 1) + 2 * ( cos x + 1))#

#=> ((cancel(1 + cos x )) * (1 - cos x)) / ((cancel(cos x +1)) (cos x + 2))#

#=> (1 - cos x) / (2 + cos x) = R H S#

Q E D

Apr 16, 2018

Kindly refer to a Proof in the Explanation.

Explanation:

#sin^2x/(cos^2x+3cosx+2)#,

#=(1-cos^2x)/{(cosx+1)(cosx+2)}#,

#={cancel((1+cosx))(1-cosx)}/{cancel((cosx+1))(cosx+2)}#,

#=(1-cosx)/(2+cosx)#, as desired!

Apr 16, 2018

#"see explanation"#

Explanation:

#"using the "color(blue)"trigonometric identity"#

#•color(white)(x)sin^2x+cos^2x=1#

#rArrsin^2x=1-cos^2x#

#"consider the left side"#

#(sin^2x)/(cos^2x+3cosx+2)#

#=(1-cos^2x)/(cos^2x+3cosx+2)#

#"the numerator is a "color(blue)"difference of squares"#

#"and the denominator is a quadratic in cos"#

#=((1-cosx)cancel((1+cosx)))/((cosx+2)cancel((cosx+1)))#

#=(1-cosx)/(2+cosx)=" right side "rArr"proven"#

Apr 16, 2018

Please see below.
After replacing :#cos^2. tocos^2x#

Explanation:

Here,

#sin^2x/(cos^2color(red)x+3cosx+2)=(1-cosx)/(2+cosx)#

Now,

#cos^2x+3cosx+2=cos^2x+2cosx+cosx+2#

#color(white)(..............................)=cosx(cosx+2)+1(cosx+2)#

#color(white)(..............................)=(cosx+1)(cosx+2)#

So,

#LHS=sin^2x/(cos^2x+3cosx+2)#

#=(1-cos^2x)/((cosx+1)(cosx+2))#

#=((1-cosx)cancel((1+cosx)))/(cancel((1+cosx))(2+cosx))#

#=(1-cosx)/(2+cosx)#

#=RHS#