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Exercise 3.1 · Q11

Q.If csc⁡θ−sin⁡θ=a3\csc\theta - \sin\theta = a^3 and sec⁡θ−cos⁡θ=b3\sec\theta - \cos\theta = b^3, then prove that a2b2(a2+b2)=1a^2b^2(a^2+b^2)=1.

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Simplifying the two given relations expresses a3sin⁡θa^3\sin\theta and b3cos⁡θb^3\cos\theta each as a trig square; combining them algebraically shows the clean facts a2b=cos⁡θa^2b=\cos\theta and ab2=sin⁡θab^2=\sin\theta, after which sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 finishes the proof directly.

Step 1. Simplify csc⁡θ−sin⁡θ\csc\theta-\sin\theta.

csc⁡θ−sin⁡θ=1sin⁡θ−sin⁡θ=1−sin⁡2θsin⁡θ=cos⁡2θsin⁡θ=a3 ⇒ a3sin⁡θ=cos⁡2θ.(⋆)\csc\theta-\sin\theta=\frac{1}{\sin\theta}-\sin\theta=\frac{1-\sin^2\theta}{\sin\theta}=\frac{\cos^2\theta}{\sin\theta}=a^3 \ \Rightarrow\ a^3\sin\theta=\cos^2\theta. \quad (\star)

Step 2. Simplify sec⁡θ−cos⁡θ\sec\theta-\cos\theta.

sec⁡θ−cos⁡θ=1cos⁡θ−cos⁡θ=1−cos⁡2θcos⁡θ=sin⁡2θcos⁡θ=b3 ⇒ b3cos⁡θ=sin⁡2θ.(⋆⋆)\sec\theta-\cos\theta=\frac{1}{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\frac{\sin^2\theta}{\cos\theta}=b^3 \ \Rightarrow\ b^3\cos\theta=\sin^2\theta. \quad (\star\star)

Step 3. Derive a2b=cos⁡θa^2b=\cos\theta. Square (⋆)(\star): a6sin⁡2θ=cos⁡4θa^6\sin^2\theta=\cos^4\theta. Substitute sin⁡2θ=b3cos⁡θ\sin^2\theta=b^3\cos\theta from (⋆⋆)(\star\star):

a6⋅b3cos⁡θ=cos⁡4θ ⇒ a6b3=cos⁡3θ ⇒ (a2b)3=cos⁡3θ ⇒ a2b=cos⁡θa^6\cdot b^3\cos\theta=\cos^4\theta \ \Rightarrow\ a^6b^3=\cos^3\theta \ \Rightarrow\ (a^2b)^3=\cos^3\theta \ \Rightarrow\ a^2b=\cos\theta

(taking real cube roots, valid since a,b,cos⁡θa,b,\cos\theta are taken as real numbers in the admissible range). …

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