Question

question given cos a = 60/61 and that angle a is in quadrant i, find the exact value of cot a in simplest radical form using a rational denominator. answer attempt 1 out of 2

Explanation:

Step1: Recall the Pythagorean identity

We know that \(\sin^{2}A+\cos^{2}A = 1\). Given \(\cos A=\frac{60}{61}\), then \(\sin^{2}A=1-\cos^{2}A\).
Substitute \(\cos A=\frac{60}{61}\) into the formula: \(\sin^{2}A=1 - (\frac{60}{61})^{2}=\frac{61^{2}-60^{2}}{61^{2}}\).
Using the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), where \(a = 61\) and \(b = 60\), we have \(61^{2}-60^{2}=(61 + 60)(61-60)=121\). So \(\sin^{2}A=\frac{121}{61^{2}}\), and since angle \(A\) is in Quadrant I (\(\sin A>0\)), \(\sin A=\frac{11}{61}\).

Step2: Recall the cotangent formula

The formula for \(\cot A=\frac{\cos A}{\sin A}\).
Substitute \(\cos A=\frac{60}{61}\) and \(\sin A=\frac{11}{61}\) into the formula: \(\cot A=\frac{\frac{60}{61}}{\frac{11}{61}}=\frac{60}{11}\).

Answer:

\(\frac{60}{11}\)