write all the other trigonometric ratios of angle A in terms of sec A

Answer: cos⁡A=1sec⁡A,sin⁡A=sec⁡2A−1sec⁡A,tan⁡A=sec⁡2A−1,cot⁡A=1sec⁡2A−1,cosec A=sec⁡Asec⁡2A−1\cos A = \frac{1}{\sec A}, \quad \sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}, \quad \tan A = \sqrt{\sec^2 A - 1}, \quad \cot A = \frac{1}{\sqrt{\sec^2 A - 1}}, \quad \text{cosec } A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}

Step-by-step solution

Step 1: Express cos⁡A\cos A in terms of sec⁡A\sec A

We know by definition that cosine is the reciprocal of secant. Therefore, we directly express cos⁡A\cos A as 1sec⁡A\frac{1}{\sec A}.

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

Using the trigonometric identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1, we have sin⁡2A=1−cos⁡2A\sin^2 A = 1 - \cos^2 A. For an acute angle AA, sin⁡A=1−cos⁡2A\sin A = \sqrt{1 - \cos^2 A}. Substituting cos⁡A=1sec⁡A\cos A = \frac{1}{\sec A}, we simplify to get sin⁡A=sec⁡2A−1sec⁡A\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}.

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

We use the identity 1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A. Rearranging for tan⁡2A\tan^2 A, we get tan⁡2A=sec⁡2A−1\tan^2 A = \sec^2 A - 1. Taking the positive square root for acute angle AA gives tan⁡A=sec⁡2A−1\tan A = \sqrt{\sec^2 A - 1}.

Step 4: Express cot⁡A\cot A in terms of sec⁡A\sec A

Since cotangent is the reciprocal of tangent, we substitute the expression found for tan⁡A\tan A. This gives cot⁡A=1sec⁡2A−1\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}.

Step 5: Express cosec A\text{cosec } A in terms of sec⁡A\sec A

Cosecant is the reciprocal of sine. Inverting our expression for sin⁡A\sin A gives cosec A=sec⁡Asec⁡2A−1\text{cosec } A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}.

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