Express all the trigonometric ratios in terms of Sin A

Answer: cos⁡A=1−sin⁡2A,cosec A=1sin⁡A,tan⁡A=sin⁡A1−sin⁡2A,sec⁡A=11−sin⁡2A,cot⁡A=1−sin⁡2Asin⁡A\cos A = \sqrt{1 - \sin^2 A}, \quad \text{cosec } A = \frac{1}{\sin A}, \quad \tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}, \quad \sec A = \frac{1}{\sqrt{1 - \sin^2 A}}, \quad \cot A = \frac{\sqrt{1 - \sin^2 A}}{\sin A}

Step-by-step solution

Step 1: Express cos⁡A\cos A using the fundamental identity

We use the fundamental identity cos⁡2A+sin⁡2A=1\cos^2 A + \sin^2 A = 1. Rearranging gives cos⁡2A=1−sin⁡2A\cos^2 A = 1 - \sin^2 A, which implies cos⁡A=±1−sin⁡2A\cos A = \pm \sqrt{1 - \sin^2 A}. For an acute angle AA, cos⁡A\cos A is positive, so we take cos⁡A=1−sin⁡2A\cos A = \sqrt{1 - \sin^2 A}.

Step 2: Express cosec A\text{cosec } A using the reciprocal relation

By definition, the cosecant ratio is the reciprocal of the sine ratio. Therefore, we directly express cosec A=1sin⁡A\text{cosec } A = \frac{1}{\sin A}.

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

The tangent of an angle is the quotient of its sine and cosine, so tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}. Substituting cos⁡A=1−sin⁡2A\cos A = \sqrt{1 - \sin^2 A} gives the ratio entirely in terms of sin⁡A\sin A.

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

The secant ratio is the reciprocal of cos⁡A\cos A. Substituting our earlier expression for cos⁡A\cos A, we obtain sec⁡A=11−sin⁡2A\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}.

Step 5: Express cot⁡A\cot A in terms of sin⁡A\sin A

The cotangent ratio is the reciprocal of tan⁡A\tan A, or alternatively cos⁡Asin⁡A\frac{\cos A}{\sin A}. Inverting our expression for tan⁡A\tan A yields cot⁡A=1−sin⁡2Asin⁡A\cot A = \frac{\sqrt{1 - \sin^2 A}}{\sin A}.

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