Express the trigonometric ratio sign a sec a tan a in terms of cot a

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

Step-by-step solution

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

We begin by expressing tan⁡A\tan A in terms of cot⁡A\cot A. By the reciprocal identity of trigonometry, the tangent of an angle is directly the reciprocal of its cotangent.

Step 2: Express sin⁡A\sin A in terms of cot⁡A\cot A

To express sin⁡A\sin A, recall the identity csc⁡2A=1+cot⁡2A\csc^2 A = 1 + \cot^2 A. Taking the positive square root for acute angle AA, we get csc⁡A=1+cot⁡2A\csc A = \sqrt{1 + \cot^2 A}. Since sin⁡A=1csc⁡A\sin A = \frac{1}{\csc A}, we obtain sin⁡A=11+cot⁡2A\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}.

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

We can express sec⁡A\sec A using 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} 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}. Taking the square root gives sec⁡A=1+cot⁡2Acot⁡A\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}.

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