Question

Question #d6cdd

Answer 1

The correct answer is `d`. See below.

Explanation:

Make sure you understand how to work with fractions. I would use the same approach as you wished to use.

We use `tanx = sinx/cosx` and `cotx = cosx/sinx`:

`=1 - sin^2x/(1 + cosx/sinx)- cos^2x/(1 +sinx/cosx)`

`=1 - sin^2x/((sinx + cosx)/sinx) - cos^2x/((cosx + sinx)/cosx)`

`=1 - ((sin^2x)sinx)/(sinx + cosx) - (cos^2x(cosx))/(cosx+ sinx)`

`=(sinx + cosx - sin^3x - cos^3x)/(cosx + sinx)`

You can factor the expression in the numerator.

`=(sinx(1 - sin^2x) + cosx(1 - cos^2x))/(cosx + sinx)`

Use `sin^2x + cos^2x = 1`.

`=(sinx(cos^2x) + cosx(sin^2x))/(cosx + sinx)`

`=(sinxcosx(cosx + sinx))/(cosx + sinx)`

`=sinxcosx`

So, the answer is `d`.

Hopefully this helps!

Answer 2

Refer to the Explanation.

Explanation:

The Expression`=1-sin^2x/(1+cotx)-cos^2x/(1+tanx)`

`=1-sin^2x/{1+cosx/sinx}-cos^2x/{1+sinx/cosx}`

`=1-sin^3x/(sinx+cosx)-cos^3x/(cosx+sinx)`

`=1-{(sin^3x+cos^3x)/(sinx+cosx)}`

Using, `a^3+b^3=(a+b)(a^2-ab+b^2)`, we get,

`"The Exp."=1-{(sinx+cosx)(sin^2x-sinxcosx+cos^2x)}/(sinx+cosx)`

`=1-{(1-sinxcosx)},.........[because, sin^2x+cos^2x=1]`

`:." The Exp.="sinxcosx, i.e., "Option d)."`

Enjoy Maths.!

Answer 3

"D" is the Correct Answer

Explanation:

first of all factor the (-1) from `-sin^2x/(1+cotx)-cos^2x/(1+tanx)`

you get `1-(sin^2x/(1+cotx)+cos^2x/(1+tanx))`

simplify each term on its own then add them

`sin^2x/(1+cotx)` = `sin^2/(1+cosx/sinx` = `sin^2/((sinx+cosx)/sinx)` = `sin^3x/(sinx+cosx)` -->"1"

`cos^2x/(1+tanx)` = `cos^2x/(1+sinx/cosx)` = `cos^2x/((sinx+cosx)/cosx)` = `cos^3x/(cosx+sinx)` --> "2"

summing 1,2

`(sin^3x+cos^3x)/(sinx+cosx)`

= `((sinx+cosx)(sin^2x-sinx*cosx+cos^2x))/(sinx+cosx)`

"cancelling `Sinx+cosx`" : `(sin^2x-sinx*cosx+cos^2x)` =

`1-cosx*sinx`

back to the full view : `1-(1-Cosx+sinx)` = `Cosx+sinx`