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Question Number 41389 by mondodotto@gmail.com last updated on 06/Aug/18

$$\mathbf{if}{\mathbf{tan}}^2\mathbf{A}+{\mathbf{tan}}^2\mathbf{B}+3=0$$$$\mathbf{show}\mathbf{that}{\mathbf{cos}}^2\mathbf{B}+2{\mathbf{cos}}^2\mathbf{A}=0$$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Aug/18

$$plscheckthequestion$$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18

$${tan}^{2}A+{tan}^{2}B+3cannotbezero$$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18

$${sec}^{2}A+{sec}^{2}B+1=0$$$${sec}^{2}B=-1-{sec}^{2}A$$$$\frac{1}{{cos}^{2}B}=-(1+\frac{1}{{cos}^{2}A})$$$$\frac{1}{{cos}^{2}B}=-\left(\frac{1+{cos}^{2}A}{{cos}^{2}A}\right)$$$${cos}^{2}B=-\left(\frac{{cos}^{2}A}{1+{cos}^{2}A}\right)$$$$so$$$${cos}^{2}B+2{cos}^{2}A$$$$=-\left(\frac{{cos}^{2}A}{1+{cos}^{2}A}\right)+2{cos}^{2}A$$$$={cos}^{2}A(2-\frac{1}{1+{cos}^{2}A})$$$$={cos}^{2}A\left(\frac{2+2{cos}^{2}A-1}{1+{cos}^{2}A}\right)$$$$={cos}^{2}A\left(\frac{1+2{cos}^{2}A}{1+{cos}^{2}A}\right)$$$$\mathit{p}\mathit{l}\mathit{s}\mathit{c}\mathit{h}\mathit{e}\mathit{c}\mathit{k}\mathit{t}\mathit{h}\mathit{e}\mathit{q}\mathit{u}\mathit{e}\mathit{s}\mathit{t}\mathit{i}\mathit{o}\mathit{n}...$$$$$$

Commented by mondodotto@gmail.com last updated on 07/Aug/18

$${\mathbf{tan}}^2\mathbf{A}+2{\mathbf{tan}}^2\mathbf{B}+3=0$$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18

$$ok$$$${tan}^{2}A+1+2(1+{tan}^{2}B)=0$$$${sec}^{2}A+2{sec}^{2}B=0$$$$\frac{1}{{cos}^{2}A}+\frac{2}{{cos}^{2}B}=0$$$${cos}^{2}B+2{cos}^{2}A=0$$

Commented by mondodotto@gmail.com last updated on 07/Aug/18

$$\mathrm{thanx}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18

$$itsok...$$

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