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Question Number 143350 by gsk2684 last updated on 13/Jun/21

$$\mathrm{if}{\tan }^2\alpha {\tan }^2\beta +{\tan }^2\beta {\tan }^2\gamma +$$$${\tan }^2\gamma {\tan }^2\alpha +{2\tan }^2\alpha {\tan }^2\beta {\tan }^2\gamma =1$$$$\mathrm{then}\mathrm{find}{\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma$$

Answered by mnjuly1970 last updated on 13/Jun/21

$$\mathrm{\Omega }:={sin}^2\alpha +{sin}^2\beta +{sin}^2\gamma$$$$:=\frac{{tan}^2\alpha }{1+{tan}^2\alpha }+\frac{{tan}^2\beta }{1+{tan}^2\beta }+\frac{{tan}^2\gamma }{1+{tan}^2\gamma }$$$$:=\frac{{tan}^2\alpha +{tan}^2\beta +{tan}^2\gamma +2{tan}^2\alpha {tan}^2\beta +2{tan}^2\alpha {tan}^2\gamma +2{tan}^2\beta {tan}^2\gamma +3{tan}^2\alpha {tan}^2\beta {tan}^2\gamma }{1+{tan}^2\alpha +{tan}^2\beta +{tan}^2\gamma +{tan}^2\alpha {tan}^2\beta +{tan}^2\beta {tan}^2\gamma +{tan}^2\alpha {tan}^2\gamma +{tan}^2\alpha {tan}^2\beta {tan}^2\gamma }$$$$:=\frac{[{tan}^2\alpha +{tan}^2\beta +{tan}^2\gamma +{tan}^2\alpha {tan}^2\beta +{tan}^2\alpha {tan}^2\gamma +{tan}^2\beta {tan}^2\gamma +{tan}^2+{tan}^2\alpha {tan}^2\beta +{tan}^2\gamma +1]:=\mathrm{A}}{\mathrm{A}}=1$$$$........\mathrm{\Omega }:=1........$$$$\{Note:={sin}^2\left(x\right)=\frac{{tan}^2\left(x\right)}{1+{tan}^2\left(x\right)}\}$$

Commented by gsk2684 last updated on 13/Jun/21

$$\mathrm{thank}\mathrm{you}$$

Commented by mnjuly1970 last updated on 13/Jun/21

$$thankyousomuchsir...$$

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