Trigonometry Practice Test 1

Concept:Simplify trigonometric fractions using identities for $\sec A$, $\csc A$, and standard factorization formulas.Explanation:Given expression:$$\frac{1-\sin A\cos A}{\cos A(\sec A-\csc A)}\cdot\frac{\sin^2 A-\cos^2 A}{\sin^3 A+\cos^3 A}$$First simplify $\sec A-\csc A$:$$\sec A-\csc A=\frac{1}{\cos A}-\frac{1}{\sin A}=\frac{\sin A-\cos A}{\sin A\cos A}$$So the denominator of the first fraction becomes:$$\cos A(\sec A-\csc A)=\cos A\cdot\frac{\sin A-\cos A}{\sin A\cos A}=\frac{\sin A-\cos A}{\sin A}$$Therefore, the first fraction is:$$\frac{1-\sin A\cos A}{\frac{\sin A-\cos A}{\sin A}}=\frac{\sin A(1-\sin A\cos A)}{\sin A-\cos A}$$Now factorize the second fraction:$$\sin^2 A-\cos^2 A=(\sin A-\cos A)(\sin A+\cos A)$$$$\sin^3 A+\cos^3 A=(\sin A+\cos A)(\sin^2 A-\sin A\cos A+\cos^2 A)$$Since $\sin^2 A+\cos^2 A=1$,$$\sin^3 A+\cos^3 A=(\sin A+\cos A)(1-\sin A\cos A)$$Thus the second fraction becomes:$$\frac{(\sin A-\cos A)(\sin A+\cos A)}{(\sin A+\cos A)(1-\sin A\cos A)}=\frac{\sin A-\cos A}{1-\sin A\cos A}$$Multiply the two simplified fractions:$$\frac{\sin A(1-\sin A\cos A)}{\sin A-\cos A}\cdot\frac{\sin A-\cos A}{1-\sin A\cos A}=\sin A$$Answer:The value is $\sin A$, so the correct option is C.