Question

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

Answer 1

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

Answer 2

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!

Answer 3

`"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"`

Answer 4

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`