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Exercise 3.1 · Q8

Q.If tan⁡2θ=1−k2\tan^2\theta = 1-k^2, show that sec⁡θ+tan⁡3θcsc⁡θ=(2−k2)3/2\sec\theta + \tan^3\theta\csc\theta = (2-k^2)^{3/2}. Also, find the values of kk for which this result holds.

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Using sec⁡2θ=1+tan⁡2θ\sec^2\theta=1+\tan^2\theta converts the given condition into sec⁡2θ=2−k2\sec^2\theta=2-k^2 directly; combining the left side over the common denominator cos⁡3θ\cos^3\theta collapses it to sec⁡3θ\sec^3\theta, matching (2−k2)3/2(2-k^2)^{3/2}. The restriction on kk comes from requiring tan⁡2θ≥0\tan^2\theta\ge0.

Step 1. Relate sec⁡2θ\sec^2\theta to kk. From sec⁡2θ=1+tan⁡2θ\sec^2\theta=1+\tan^2\theta and the given tan⁡2θ=1−k2\tan^2\theta=1-k^2:

sec⁡2θ=1+(1−k2)=2−k2.\sec^2\theta=1+(1-k^2)=2-k^2.

Step 2. Rewrite the left side over a common denominator.

sec⁡θ+tan⁡3θcsc⁡θ=1cos⁡θ+sin⁡3θcos⁡3θ⋅1sin⁡θ=1cos⁡θ+sin⁡2θcos⁡3θ=cos⁡2θ+sin⁡2θcos⁡3θ.\sec\theta+\tan^3\theta\csc\theta = \frac{1}{\cos\theta}+\frac{\sin^3\theta}{\cos^3\theta}\cdot\frac{1}{\sin\theta}=\frac{1}{\cos\theta}+\frac{\sin^2\theta}{\cos^3\theta}=\frac{\cos^2\theta+\sin^2\theta}{\cos^3\theta}.

Step 3. Simplify using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1.

cos⁡2θ+sin⁡2θcos⁡3θ=1cos⁡3θ=sec⁡3θ.\frac{\cos^2\theta+\sin^2\theta}{\cos^3\theta}=\frac{1}{\cos^3\theta}=\sec^3\theta.

Step 4. Substitute sec⁡2θ=2−k2\sec^2\theta=2-k^2 from Step 1.

sec⁡3θ=(sec⁡2θ)3/2=(2−k2)3/2.\sec^3\theta=(\sec^2\theta)^{3/2}=(2-k^2)^{3/2}. …

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