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Question Number 25536 by tawa tawa last updated on 11/Dec/17

If   x = 2sec(3t)  and  y = 4tan(3t),   Show that    y = 4(x^2  − 4)

$$\mathrm{If}\:\:\:\mathrm{x}\:=\:\mathrm{2sec}\left(\mathrm{3t}\right)\:\:\mathrm{and}\:\:\mathrm{y}\:=\:\mathrm{4tan}\left(\mathrm{3t}\right),\:\:\:\mathrm{Show}\:\mathrm{that}\:\:\:\:\mathrm{y}\:=\:\mathrm{4}\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4}\right) \\ $$

Answered by prakash jain last updated on 11/Dec/17

x=2sec 3t⇒(x/2)=sec 3t  y=4tan 3t⇒(y/4)=tan 3t  sec^2 θ−tan^2 θ=1  (x^2 /4)−(y^2 /(16))=1  ((x^2 −4)/4)=(y^2 /(16))  ⇒y^2 =4(x^2 −4)

$${x}=\mathrm{2sec}\:\mathrm{3}{t}\Rightarrow\frac{{x}}{\mathrm{2}}=\mathrm{sec}\:\mathrm{3}{t} \\ $$$${y}=\mathrm{4tan}\:\mathrm{3}{t}\Rightarrow\frac{{y}}{\mathrm{4}}=\mathrm{tan}\:\mathrm{3}{t} \\ $$$$\mathrm{sec}^{\mathrm{2}} \theta−\mathrm{tan}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{{y}^{\mathrm{2}} }{\mathrm{16}}=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4}}=\frac{{y}^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\Rightarrow{y}^{\mathrm{2}} =\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{4}\right) \\ $$

Commented by tawa tawa last updated on 13/Dec/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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