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

trig 
Reiny,
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/(sinxcosx))  ((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 
trig 
Evelina,
The b) actually was like this in my book: sin^2x/sinxcosx  sinx+cosx/tg^2x+1 = sinx + cosx

trig 
Reiny,
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/sinxcosx  sinx+cosx/tg^2x+1 = sinx + cosx and LS ≠ RS
I tried it in
sin^2x/(sinxcosx)  (sinx+cosx)/(tg^2x+1) = sinx + cosx and LS ≠ RS
I tried it in
sin^2x/(sinxcosx)  (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/(sinxcosx)  (sinx + cosx)/(tan^2x + 1) = sinx + cosx
and if I test x=20°, LS ≠ RS 
trig 
Evelina,
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.