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Exercise 3.1 · Q10

Q.If cot⁡θ(1+sin⁡θ)=4m\cot\theta(1+\sin\theta)=4m and cot⁡θ(1−sin⁡θ)=4n\cot\theta(1-\sin\theta)=4n, then prove that (m2−n2)2=mn(m^2-n^2)^2=mn.

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Adding and subtracting the two given relations isolates m+nm+n and m−nm-n separately in terms of θ\theta; then both (m2−n2)2=[(m−n)(m+n)]2(m^2-n^2)^2=[(m-n)(m+n)]^2 and mn=(m+n)2−(m−n)24mn=\tfrac{(m+n)^2-(m-n)^2}{4} reduce to the same expression cos⁡4θ16sin⁡2θ\dfrac{\cos^4\theta}{16\sin^2\theta}.

Step 1. Subtract the two given relations. cot⁡θ(1+sin⁡θ)−cot⁡θ(1−sin⁡θ)=4m−4n\cot\theta(1+\sin\theta)-\cot\theta(1-\sin\theta)=4m-4n

cot⁡θ⋅2sin⁡θ=4(m−n) ⇒ 2cos⁡θ=4(m−n) ⇒ m−n=cos⁡θ2\cot\theta\cdot2\sin\theta=4(m-n) \ \Rightarrow\ 2\cos\theta=4(m-n) \ \Rightarrow\ m-n=\frac{\cos\theta}{2}

(using cot⁡θsin⁡θ=cos⁡θ\cot\theta\sin\theta=\cos\theta).

Step 2. Add the two given relations. cot⁡θ(1+sin⁡θ)+cot⁡θ(1−sin⁡θ)=4m+4n\cot\theta(1+\sin\theta)+\cot\theta(1-\sin\theta)=4m+4n

cot⁡θ⋅2=4(m+n) ⇒ m+n=cot⁡θ2.\cot\theta\cdot2=4(m+n) \ \Rightarrow\ m+n=\frac{\cot\theta}{2}.

Step 3. Compute m2−n2m^2-n^2.

m2−n2=(m−n)(m+n)=cos⁡θ2⋅cot⁡θ2=cos⁡θcot⁡θ4=cos⁡2θ4sin⁡θm^2-n^2=(m-n)(m+n)=\frac{\cos\theta}{2}\cdot\frac{\cot\theta}{2}=\frac{\cos\theta\cot\theta}{4}=\frac{\cos^2\theta}{4\sin\theta}

(using cot⁡θ=cos⁡θ/sin⁡θ\cot\theta=\cos\theta/\sin\theta). So (m2−n2)2=cos⁡4θ16sin⁡2θ(m^2-n^2)^2=\dfrac{\cos^4\theta}{16\sin^2\theta}.

Step 4. Compute mnmn. From Steps 1–2, m=(m+n)+(m−n)2=cot⁡θ+cos⁡θ4m=\dfrac{(m+n)+(m-n)}{2}=\dfrac{\cot\theta+\cos\theta}{4}, and similarly n=cot⁡θ−cos⁡θ4n=\dfrac{\cot\theta-\cos\theta}{4}. So …

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