You need to evaluate `bsqrt(1-a^2)-asqrt(1-b^2)` using the information provides by the problem that `a = sin 28^o` and `b = cos 32^o` , such that:

`cos 32^o*sqrt(1 - sin^2 28^o) - sin 28^o*sqrt(1 - cos^2 32^o)`

Using the Pyhtagorean identities `1 - sin^2 28^o = cos^2 28^o` and `1 -...

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You need to evaluate `bsqrt(1-a^2)-asqrt(1-b^2)` using the information provides by the problem that `a = sin 28^o` and `b = cos 32^o` , such that:

`cos 32^o*sqrt(1 - sin^2 28^o) - sin 28^o*sqrt(1 - cos^2 32^o)`

Using the Pyhtagorean identities `1 - sin^2 28^o = cos^2 28^o` and `1 - cos^2 32^o = sin^2 32^o` yields:

`cos 32^o*sqrt(cos^2 28^o) - sin 28^o*sqrt(sin^2 32^o)`

Taking the square roots yields:

`cos 32^o*cos 28^o - sin 28^o*sin 32^o`

Using the trigonometric identity `cos alpha*cos beta - sin alpha*sin beta = cos(alpha + beta)` yields:

`cos 32^o*cos 28^o - sin 28^o*sin 32^o = cos (32^o + 28^o)`

`cos 32^o*cos 28^o - sin 28^o*sin 32^o = cos 60^o = 1/2`

**Hence, evaluating the given expression, under the given conditions, yields `bsqrt(1-a^2)-asqrt(1-b^2) = 1/2` .**

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