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b. 1-cos x=sin ^{2} x 5. Confirm the identities (9 points each) (\csc x-t x)(\csc x+t x)=1 cos x(\csc x-sec x)=t x-1

User Koichi
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To confirm the identity (1 - cos(x)) = sin^2(x) and (csc(x) - tan(x))(csc(x) + tan(x)) = 1, we can simplify each expression separately:

1. Confirming the identity (1 - cos(x)) = sin^2(x):

Starting with the left-hand side (LHS):

LHS = 1 - cos(x)

Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can rewrite the right-hand side (RHS):

sin^2(x) = 1 - cos^2(x)

Now, let's compare the LHS and RHS:

LHS = 1 - cos(x)

RHS = sin^2(x) = 1 - cos^2(x)

Since both the LHS and RHS are equal, we have confirmed the identity (1 - cos(x)) = sin^2(x).

2. Confirming the identity (csc(x) - tan(x))(csc(x) + tan(x)) = 1:

Starting with the left-hand side (LHS):

LHS = (csc(x) - tan(x))(csc(x) + tan(x))

To simplify, we'll use the reciprocal and quotient identities:

csc(x) = 1 / sin(x)

tan(x) = sin(x) / cos(x)

Substituting these identities into the expression:

LHS = (1 / sin(x) - sin(x) / cos(x))(1 / sin(x) + sin(x) / cos(x))

To simplify further, we can find a common denominator:

LHS = [(cos(x) - sin^2(x)) / (sin(x) * cos(x))][(cos(x) + sin^2(x)) / (sin(x) * cos(x))]

Now, let's multiply the numerators:

LHS = (cos^2(x) - sin^4(x)) / (sin^2(x) * cos^2(x))

Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can rewrite the numerator:

LHS = (cos^2(x) - (1 - cos^2(x))^2) / (sin^2(x) * cos^2(x))

LHS = (cos^2(x) - (1 - 2cos^2(x) + cos^4(x))) / (sin^2(x) * cos^2(x))

LHS = (3cos^2(x) - cos^4(x) - 1) / (sin^2(x) * cos^2(x))

Simplifying further, we have:

LHS = (3cos^2(x) - cos^4(x) - 1) / (sin^2(x) * cos^2(x))

Since the expression does not simplify to 1, the identity (csc(x) - tan(x))(csc(x) + tan(x)) = 1 does not hold true.

Therefore, only the first identity (1 - cos(x)) = sin^2(x) is confirmed, while the second identity is not valid.

User Meshaal
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