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Question Number 87027 by john santu last updated on 02/Apr/20

Answered by som(math1967) last updated on 05/Apr/20

$$\mathrm{\angle }DCO=alt\mathrm{\angle }OAP=\theta \left(let\right)$$$$\mathrm{\angle }COQ=\mathrm{\angle }OAP=\theta$$$$again\mathrm{\angle }ADO=\mathrm{\angle }DCO=\theta ns$$$$[△ADO\sim △DCO]$$$$OC=nsec\theta ,OA=OPcosec\theta =cosec\theta$$$$[\because n=OQ]$$$$\therefore OC+OA=AC$$$$nsec\theta +cosec\theta =d.......1)$$$$nowOD=OCtan\theta [from△ODC]$$$$OD=nsec\theta tan\theta$$$$from△ODAOD=OAcot\theta$$$$\therefore OD=cosec\theta cot\theta$$$$nsec\theta tan\theta =cosec\theta cot\theta$$$${ntan}^2\theta =cot\theta$$$${tan}^3\theta =\frac{1}{n}tan\theta ={\left(\frac{1}{n}\right)}^{\frac{1}{3}}$$$$\therefore sec\theta =\frac{{(1+n^{\frac{2}{3}})}^{\frac{1}{2}}}{n^{\frac{1}{3}}}$$$$cosec\theta =\frac{{(1+n^{\frac{2}{3}})}^{\frac{1}{2}}}{1}$$$$from1)nsec\theta +cosec\theta =d$$$$n.\frac{{(1+n^{\frac{2}{3}})}^{\frac{1}{2}}}{n^{\frac{1}{3}}}+{(1+n^{\frac{2}{3}})}^{\frac{1}{2}}=d$$$$n^{\frac{2}{3}}{(1+n^{\frac{2}{3}})}^{\frac{1}{2}}+{(1+n^{\frac{2}{3}})}^{\frac{1}{2}}=d$$$${(1+n^{\frac{2}{3}})}^{\frac{1}{2}}(1+n^{\frac{2}{3}})=d$$$${(1+n^{\frac{2}{3}})}^{\frac{3}{2}}=d$$$$\therefore 1+n^{\frac{2}{3}}=d^{\frac{2}{3}}$$$$$$

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