Mathematics

Mathematics

Problem

61. Multiple Choice Which of these is an efficient first step in proving the identity sin⁡x1−cos⁡x=1+cos⁡xsin⁡x\frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x} ? (A) sin⁡x1−cos⁡x=cos⁡(π2−x)1−cos⁡x\frac{\sin x}{1-\cos x}=\frac{\cos \left(\frac{\pi}{2}-x\right)}{1-\cos x} (B) sin⁡x1−cos⁡x=sin⁡xsin⁡2x+cos⁡2x−cos⁡x\frac{\sin x}{1-\cos x}=\frac{\sin x}{\sin ^{2} x+\cos ^{2} x-\cos x} (C) sin⁡x1−cos⁡x=sin⁡x1−cos⁡x⋅csc⁡xcsc⁡x\frac{\sin x}{1-\cos x}=\frac{\sin x}{1-\cos x} \cdot \frac{\csc x}{\csc x} (D) sin⁡x1−cos⁡x=sin⁡x1−cos⁡x⋅1−cos⁡x1−cos⁡x\frac{\sin x}{1-\cos x}=\frac{\sin x}{1-\cos x} \cdot \frac{1-\cos x}{1-\cos x} (E) sin⁡x1−cos⁡x=sin⁡x1−cos⁡x⋅1+cos⁡x1+cos⁡x\frac{\sin x}{1-\cos x}=\frac{\sin x}{1-\cos x} \cdot \frac{1+\cos x}{1+\cos x} (A) sin⁡x1−cos⁡x=cos⁡(π2−x)1−cos⁡x\frac{\sin x}{1-\cos x}=\frac{\cos \left(\frac{\pi}{2}-x\right)}{1-\cos x} (B) sin⁡x1−cos⁡x=sin⁡xsin⁡2x+cos⁡2x−cos⁡x\frac{\sin x}{1-\cos x}=\frac{\sin x}{\sin ^{2} x+\cos ^{2} x-\cos x} (C) sin⁡x1−cos⁡x=sin⁡x1−cos⁡x⋅csc⁡xcsc⁡x\frac{\sin x}{1-\cos x}=\frac{\sin x}{1-\cos x} \cdot \frac{\csc x}{\csc x} (D) sin⁡x1−cos⁡x=sin⁡x1−cos⁡x⋅1−cos⁡x1−cos⁡x\frac{\sin x}{1-\cos x}=\frac{\sin x}{1-\cos x} \cdot \frac{1-\cos x}{1-\cos x} (E) sin⁡x1−cos⁡x=sin⁡x1−cos⁡x⋅1+cos⁡x1+cos⁡x\frac{\sin x}{1-\cos x}=\frac{\sin x}{1-\cos x} \cdot \frac{1+\cos x}{1+\cos x}

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