Question

I need to prove equality.
a) (sina + cosa)^2 -1 / ctga - sinacosa = 2tg^2a
b) (sin^2x/sinx-cosx) - (sinx+cosx/tg^2x+1) = sinx + cosx
c) sin^4a - sin^2a - cos^4a + cos^2a = cosπ/2
How to do these?

Answer 1

a)
LS = [sin^2a + 2sinacosa + cos^2a - 1] / [ cosa/sina - sinacosa]
= [1 + 2sinacosa -1] / [ (cosa - sin^2acosa)/sina]
= 2sinacosa/[ cosa(1 - sin^2a)/sina]
= 2sinacosa(sina/(cosa(cos^2a))
= 2 sin^2a/cos^2a
= 2tan^2a
= RS

b) The way you typed it, LS ≠ RS
Did you mean
(sin^2x/(sinx-cosx)) - ((sinx+cosx)/tg^2x+1) = sinx + cosx ?
Please clarify

c) LS = sin^2(sin^2a - 1) - cos^2a(cos^2a -1)
= sin^2a(-cos^2a) - cos^2a(-sin^2a)
= 0
RS = cos π/2
= 0
= LS

Answer 2

The b) actually was like this in my book: sin^2x/sinx-cosx - sinx+cosx/tg^2x+1 = sinx + cosx

Answer 3

When I can't seem to get anywhere with an identity I take any value of the variable and test it in the equation.
I tried x = 20° in
sin^2x/sinx-cosx - sinx+cosx/tg^2x+1 = sinx + cosx and LS ≠ RS
I tried it in
sin^2x/(sinx-cosx) - (sinx+cosx)/(tg^2x+1) = sinx + cosx and LS ≠ RS
I tried it in
sin^2x/(sinx-cosx) - (sinx+cosx)/tg^2x+1 = sinx + cosx and LS ≠ RS

You do realize that you must put brackets in this way of typing to identify which is the numerator and which is the denominator.

the way you typed it, the LS would have 5 terms
[sin^2x/sinx] - [cosx] - [sinx] + [cosx/tan^2x] + 1
I am pretty sure that is not what the question says.

I have a feeling there are two fractions
numerator of 1st fraction : sin^2x
denominator of 1st fraction: sinx - cosx

numerator of 2nd : sinx + cosx
denom of 2nd : tan^2x + 1
Thus

sin^2x/(sinx-cosx) - (sinx + cosx)/(tan^2x + 1) = sinx + cosx
and if I test x=20°, LS ≠ RS

Answer 4

Well, that‘s what the question says. I can send a picture of it, if you don‘t believe.
Show me how you do it with brackets, maybe I‘ll know what to do with that one.