The picture in the attached figure

we know that
The Intersecting Secants Theorem, states that
''If two secant segments are drawn to a circle from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment''
so
in this problem
(x+DC)*DC=(AB+BC)*BC
AB=42
BC=18
CD=4
(x+4)*4=(42+18)*18
4x+16=1080
4x=1080-16
x=1064/4
x=266

the answer is
x=266

5 0

3 years ago

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Cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3(x) prove the trig identity

adelina 88 [10]

Working with the right side:

cot(x) + 2 tan(x) + tan³(x) = cos(x)/sin(x) + 2 sin(x)/cos(x) + sin³(x)/cos³(x)

… = (cos⁴(x) + 2 sin²(x) cos²(x) + sin⁴(x)) / (sin(x) cos³(x))

Factorize the numerator as a sum of squares:

a⁴ + 2 a² b² + b⁴ = (a² + b²)²

… = (cos²(x) + sin²(x))² / (sin(x) cos³(x))

Recall that

cos²(x) + sin²(x) = 1

… = 1 / (sin(x) cos³(x))

… = 1 / (sin(x) cos³(x)) • cos(x)/cos(x)

… = cos(x) / (sin(x) cos⁴(x))

… = cot(x) sec⁴(x)

7 0

3 years ago