Water has a freezing point of 32 degrees. Murcury has a freezing point that is 70 degrees lower. What is the freezing point of m
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1+tan^2(A) = sec^2(A) [Pythagorean Identities]
tan^2(A)cot(A) = tan(A)[tan(A)cot(A)] = tan(A)[1] = tan(A)
*see photo for complete solution*
tan^2(A)cot(A) = tan(A)[tan(A)cot(A)] = tan(A)[1] = tan(A)
*see photo for complete solution*