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Question-45281




Question Number 45281 by peter frank last updated on 11/Oct/18
Commented by math1967 last updated on 11/Oct/18
pls chq. q.no1
$${pls}\:{chq}.\:{q}.{no}\mathrm{1} \\ $$
Commented by peter frank last updated on 11/Oct/18
i forgot this“=0” very sorry
$$\mathrm{i}\:\mathrm{forgot}\:\mathrm{this}“=\mathrm{0}''\:\mathrm{very}\:\mathrm{sorry} \\ $$$$ \\ $$
Answered by math1967 last updated on 11/Oct/18
2) siny=2sinx  sin^2 y=4sin^2 x  ∴4−sin^2 y=4−4sin^2 x  3+cos^2 y=4cos^2 x  now siny=2sinx  cosy(dy/dx)=2cosx  ((dy/dx))^2 =((4cos^2 x)/(cos^2 y))=((3+cos^2 y)/(cos^2 y))=3sec^2 y+1
$$\left.\mathrm{2}\right)\:{siny}=\mathrm{2}{sinx} \\ $$$${sin}^{\mathrm{2}} {y}=\mathrm{4}{sin}^{\mathrm{2}} {x} \\ $$$$\therefore\mathrm{4}−{sin}^{\mathrm{2}} {y}=\mathrm{4}−\mathrm{4}{sin}^{\mathrm{2}} {x} \\ $$$$\mathrm{3}+{cos}^{\mathrm{2}} {y}=\mathrm{4}{cos}^{\mathrm{2}} {x} \\ $$$${now}\:{siny}=\mathrm{2}{sinx} \\ $$$${cosy}\frac{{dy}}{{dx}}=\mathrm{2}{cosx} \\ $$$$\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\frac{\mathrm{4}{cos}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {y}}=\frac{\mathrm{3}+{cos}^{\mathrm{2}} {y}}{{cos}^{\mathrm{2}} {y}}=\mathrm{3}{sec}^{\mathrm{2}} {y}+\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by peter frank last updated on 11/Oct/18
thank you  sir.pls help no 1
$$\mathrm{thank}\:\mathrm{you}\:\:\mathrm{sir}.\mathrm{pls}\:\mathrm{help}\:\mathrm{no}\:\mathrm{1} \\ $$

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