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EXERCISE 1.4 · Q152

Q.Find dydx\dfrac{dy}{dx} if x=acot⁡θx=a\cot\theta, y=b cosec θy=b\,\text{cosec}\,\theta.

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✓ Free question

We have x=acot⁡θx=a\cot\theta and y=b cosec θy=b\,\text{cosec}\,\theta.

Step 1. dxdθ=−a cosec2θ\dfrac{dx}{d\theta}=-a\,\text{cosec}^2\theta

Step 2. dydθ=−b cosec θcot⁡θ\dfrac{dy}{d\theta}=-b\,\text{cosec}\,\theta\cot\theta

Step 3.

dydx=−b cosec θcot⁡θ−a cosec2θ=bcot⁡θa cosec θ\frac{dy}{dx}=\frac{-b\,\text{cosec}\,\theta\cot\theta}{-a\,\text{cosec}^2\theta}=\frac{b\cot\theta}{a\,\text{cosec}\,\theta}

Step 4. Since cot⁡θcosec θ=cos⁡θ/sin⁡θ1/sin⁡θ=cos⁡θ\dfrac{\cot\theta}{\text{cosec}\,\theta}=\dfrac{\cos\theta/\sin\theta}{1/\sin\theta}=\cos\theta,

dydx=bacos⁡θ\frac{dy}{dx}=\frac{b}{a}\cos\theta

✓Final answer

dydx=bacos⁡θ\dfrac{dy}{dx}=\dfrac{b}{a}\cos\theta

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