How to prove this question with using a LS=RS-style Proof?

csc^2(x)cos^2(x) = csc^2(x)-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)

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

@Don & Reiny: thank you very much..

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

To prove the equation csc^2(x) * cos^2(x) = csc^2(x) - 1 using a LS=RS (left side equals right side) proof, we need to simplify both sides of the equation and show that they are equal.

Starting with the left side (LS):
LS = csc^2(x) * cos^2(x)

Now, let's simplify the left side by using the trigonometric identity: sin^2(x) + cos^2(x) = 1. Rearranging this identity, we get: cos^2(x) = 1 - sin^2(x).

Substituting this into the left side equation, we have:
LS = csc^2(x) * (1 - sin^2(x))

Next, we need to simplify the right side (RS):
RS = csc^2(x) - 1

Now, let's simplify the right side by using the reciprocal identity: csc(x) = 1/sin(x). Squaring both sides of this identity, we get: csc^2(x) = 1/sin^2(x).

Substituting this into the right side equation, we have:
RS = 1/sin^2(x) - 1

To make the denominators the same, we will multiply the second term on the right side by sin^2(x)/sin^2(x):
RS = 1/sin^2(x) - sin^2(x)/sin^2(x)

Combining the terms:
RS = (1 - sin^2(x))/sin^2(x)

Now, we can see that both the left side (LS) and the right side (RS) are in a similar form. If we can show that they are equal, the equation will be proved.

To simplify both sides further, let's use the pythagorean identity: sin^2(x) + cos^2(x) = 1. Rearranging it, we get: sin^2(x) = 1 - cos^2(x).

Substituting this into the right side equation:
RS = (1 - (1 - cos^2(x)))/sin^2(x)
RS = (1 - 1 + cos^2(x))/sin^2(x)
RS = cos^2(x)/sin^2(x)

Now, let's simplify the left side:
LS = csc^2(x) * (1 - sin^2(x))
LS = (1/sin^2(x)) * (1 - (1 - cos^2(x)))
LS = (1/sin^2(x)) * (cos^2(x))
LS = cos^2(x)/sin^2(x)

As you can see, the left side (LS) and right side (RS) are equal. Therefore, we have proven the equation: csc^2(x) * cos^2(x) = csc^2(x) - 1 using a LS=RS-style proof.