Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
\( \sin A = \dfrac{1}{\sqrt{1 + \cot^2 A}} \)
\( \tan A = \dfrac{1}{\cot A} \)
\( \sec A = \dfrac{\sqrt{1 + \cot^2 A}}{\cot A} \)
Goal: Express \(\sin A\), \(\sec A\), and \(\tan A\) only in terms of \(\cot A\).
Step 1: Start from the basic definitions
We know the reciprocal and quotient relations:
\(\tan A = \dfrac{1}{\cot A}\) and \(\cot A = \dfrac{\cos A}{\sin A}\).
Step 2: Write \(\tan A\) in terms of \(\cot A\)
Since \(\tan A\) and \(\cot A\) are reciprocals,
\(\tan A = \dfrac{1}{\cot A}\).
Step 3: Use the identity \(1+\cot^2 A = \text{cosec }^2 A\)
One important Pythagorean identity is:
\(1 + \cot^2 A = \text{cosec }^2 A\).
Taking square root on both sides (for acute angles, values are positive):
\(\text{cosec } A = \sqrt{1 + \cot^2 A}\).
Step 4: Convert \(\text{cosec } A\) to \(\sin A\)
We know \(\text{cosec } A = \dfrac{1}{\sin A}\).
So,
\(\dfrac{1}{\sin A} = \sqrt{1 + \cot^2 A}\)
Taking reciprocal:
\(\sin A = \dfrac{1}{\sqrt{1 + \cot^2 A}}\).
Step 5: Find \(\sec A\) in terms of \(\cot A\)
We use another identity:
\(\text{cosec }^2 A = 1 + \cot^2 A\) and also \(\text{cosec }^2 A = 1 + \cot^2 A = \left(\dfrac{\text{cosec } A}{1}\right)^2\).
A quicker way is to use:
\(\cot A = \dfrac{\cos A}{\sin A}\Rightarrow \cos A = \cot A\cdot \sin A\).
Now substitute \(\sin A = \dfrac{1}{\sqrt{1+\cot^2 A}}\):
\(\cos A = \cot A \cdot \dfrac{1}{\sqrt{1+\cot^2 A}} = \dfrac{\cot A}{\sqrt{1+\cot^2 A}}\).
Since \(\sec A = \dfrac{1}{\cos A}\),
\(\sec A = \dfrac{1}{\dfrac{\cot A}{\sqrt{1+\cot^2 A}}} = \dfrac{\sqrt{1+\cot^2 A}}{\cot A}\).
Final Results:
\(\sin A = \dfrac{1}{\sqrt{1 + \cot^2 A}}\), \(\tan A = \dfrac{1}{\cot A}\), \(\sec A = \dfrac{\sqrt{1 + \cot^2 A}}{\cot A}\).
Student Tip: When you are asked to write ratios in terms of \(\cot A\), the identity \(1+\cot^2 A=\text{cosec }^2 A\) is the fastest starting point.