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Question Number 143913 by Sravanth last updated on 19/Jun/21

$$\mathrm{If}\frac{\cos \alpha }{cos\beta }=\mathrm{m}\mathrm{and}\frac{\cos \alpha }{\sin \beta }=\mathrm{n}\mathrm{then}$$$$\mathrm{prove}\mathrm{that}({\mathrm{m}}^2+{\mathrm{n}}^2){\cos }^2\beta =n^2$$

Answered by Ar Brandon last updated on 19/Jun/21

$${\cos }^2\beta +{\sin }^2\beta =1$$$${\left(\frac{\cos \alpha }{\mathrm{m}}\right)}^2+{\left(\frac{\cos \alpha }{\mathrm{n}}\right)}^2=1$$$$\frac{({\mathrm{n}}^2+{\mathrm{m}}^2){\cos }^2\alpha }{{\mathrm{m}}^2{\mathrm{n}}^2}=1$$$$\frac{({\mathrm{m}}^2+{\mathrm{n}}^2)}{{\mathrm{m}}^2{\mathrm{n}}^2}\cdot \frac{{\mathrm{m}}^2{\cos }^2\beta }{1}=1$$$$({\mathrm{m}}^2+{\mathrm{n}}^2){\cos }^2\beta ={\mathrm{n}}^2$$

Commented by Dwaipayan Shikari last updated on 19/Jun/21

$$IknowwhoisAIRBRANDONbuti{don}^{\prime }t$$$$whoisARBRANDON.ThenCongratulations$$$$too!Ityourfirstpost\left(answer\right)$$

Commented by Dwaipayan Shikari last updated on 19/Jun/21

$$IpronouceARBrandonasAirBrandon$$

Commented by Ar Brandon last updated on 19/Jun/21

$$\mathrm{Hahaha}!\mathrm{accepted}.$$$$\mathrm{AR}\to \mathrm{Arung}$$$$\mathrm{Ar}\mathrm{Brandon}\to \mathrm{Arung}\mathrm{Brandon}$$😉

Commented by Dwaipayan Shikari last updated on 19/Jun/21

$$WhatdoesArungmean?$$

Commented by Ar Brandon last updated on 19/Jun/21

It's my name ����������

Commented by Dwaipayan Shikari last updated on 19/Jun/21

$$Yesbutitmusthaveameaning$$$$LikeDwaipayanmeans‘‘Amanwho$$$$wasborninan{Island}^{″}andthewriter$$$$ofIndianEpicMaha\stackrel{-}{bharata}$$

Commented by Sravanth last updated on 19/Jun/21

Thanks Bro

Commented by Ar Brandon last updated on 19/Jun/21

It's a pleasure answering your first post. Welcome to forum Sravanth ! ��

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