This type of problem involves a lot of trig idenities, I just looked them up..
Lets look at the LHS:
csc^2(x)= 1/sin^2(x)
cos^2(x) = 1-sin^2(x)
Now we can say:
csc^2(x)*cos^2(x) = (1-sin^2(x))/sin^2(x)
now lets look at the RHS:
csc^2(x) = 1/sin^2(x)
and we can change that 1 to sin^2(x)/sin^2(x) to have same denominator
1/sin^2(x) - sin^2(x)/sin^2(x) = (1-sin^2(x))/sin^2(x)
So now LHS=RHS
(1-sin^2(x))/sin^2(x) = (1-sin^2(x))/sin^2(x)
How to prove this question with using a LS=RS-style Proof?
csc^2(x)cos^2(x) = csc^2(x)-1
3 answers
From the identity sin^2 x + cos^2 x = 1
if we divide both sides by sin^2 x , we get
1 + cot^2 x = csc^2 x
so csc^2 x - 1 = cot^2 x
RS = cot^2 x
LS = csc^2(x)cos^2(x)
= (1/sin^2 x)(cos^2 x)
= cot^2 x
= RS
All done
if we divide both sides by sin^2 x , we get
1 + cot^2 x = csc^2 x
so csc^2 x - 1 = cot^2 x
RS = cot^2 x
LS = csc^2(x)cos^2(x)
= (1/sin^2 x)(cos^2 x)
= cot^2 x
= RS
All done
@Don & Reiny: thank you very much..
I really understand your answers after 3 hours of tried them by myself. thank you so much.. :D
I really understand your answers after 3 hours of tried them by myself. thank you so much.. :D