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EXERCISE 1.4 · Q159

Q.Find dydx\dfrac{dy}{dx} if x=cosec2θx=\text{cosec}^2\theta, y=cot⁡3θy=\cot^3\theta, at θ=π6\theta=\dfrac{\pi}{6}.

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We have x=cosec2θx=\text{cosec}^2\theta, y=cot⁡3θy=\cot^3\theta.

Step 1.

dxdθ=2 cosec θ⋅(−cosec θcot⁡θ)=−2 cosec2θcot⁡θ\frac{dx}{d\theta}=2\,\text{cosec}\,\theta\cdot(-\text{cosec}\,\theta\cot\theta)=-2\,\text{cosec}^2\theta\cot\theta

Step 2.

dydθ=3cot⁡2θ⋅(−cosec2θ)=−3cot⁡2θ cosec2θ\frac{dy}{d\theta}=3\cot^2\theta\cdot(-\text{cosec}^2\theta)=-3\cot^2\theta\,\text{cosec}^2\theta

Step 3.

dydx=−3cot⁡2θ cosec2θ−2 cosec2θcot⁡θ=32cot⁡θ\frac{dy}{dx}=\frac{-3\cot^2\theta\,\text{cosec}^2\theta}{-2\,\text{cosec}^2\theta\cot\theta}=\frac32\cot\theta …

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