**EXERCISE 8.3** **1.** Express the trigonometric ratios sin⁡A\sin A, tan⁡A\tan A and sec⁡A\sec A in terms of cot⁡A\cot A.

Answer: sin⁡A=11+cot⁡2A,tan⁡A=1cot⁡A,sec⁡A=1+cot⁡2Acot⁡A\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}, \quad \tan A = \frac{1}{\cot A}, \quad \sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}

Step-by-step solution

Step 1: Express tan⁡A\tan A in terms of cot⁡A\cot A

By definition, the tangent and cotangent functions are reciprocals of each other for any acute angle AA. Therefore, tan⁡A\tan A can be written directly as 1cot⁡A\frac{1}{\cot A}.

Step 2: Express sin⁡A\sin A using the cosecant identity

We use the standard trigonometric identity cosec2A=1+cot⁡2A\text{cosec}^2 A = 1 + \cot^2 A. Since AA is an acute angle in ΔABC\Delta ABC, cosec A\text{cosec } A is positive, giving cosec A=1+cot⁡2A\text{cosec } A = \sqrt{1 + \cot^2 A}.

Step 3: Express sin⁡A\sin A in terms of cot⁡A\cot A

Since sin⁡A\sin A is the reciprocal of cosec A\text{cosec } A, we substitute cosec A=1+cot⁡2A\text{cosec } A = \sqrt{1 + \cot^2 A} to express sin⁡A\sin A in terms of cot⁡A\cot A.

Step 4: Express sec⁡A\sec A using the tangent identity

We use the identity sec⁡2A=1+tan⁡2A\sec^2 A = 1 + \tan^2 A. Substituting tan⁡A=1cot⁡A\tan A = \frac{1}{\cot A} into this identity gives sec⁡2A=1+1cot⁡2A=cot⁡2A+1cot⁡2A\sec^2 A = 1 + \frac{1}{\cot^2 A} = \frac{\cot^2 A + 1}{\cot^2 A}.

Step 5: Take the square root to find sec⁡A\sec A

Taking the square root on both sides for the acute angle AA, we obtain sec⁡A=1+cot⁡2Acot⁡A\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}.

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