Question

How to prove this question with using a LS=RS-style Proof?
csc^2(x)cos^2(x) = csc^2(x)-1

Answer 1

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)

Answer 2

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

Answer 3

@Don & Reiny: thank you very much..
I really understand your answers after 3 hours of tried them by myself. thank you so much.. :D