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Question Number 138302 by mathdave last updated on 12/Apr/21
if  tan^2 x=1+2tan^2 y  show that  cos2x+sin^2 y=0
$${if}\:\:\mathrm{tan}^{\mathrm{2}} {x}=\mathrm{1}+\mathrm{2tan}^{\mathrm{2}} {y} \\ $$$${show}\:{that} \\ $$$$\mathrm{cos2}{x}+\mathrm{sin}^{\mathrm{2}} {y}=\mathrm{0} \\ $$
Answered by Ñï= last updated on 12/Apr/21
tan^2 x=1+2tan^2 y  ⇒sec^2 x−1=2sec^2 y−1       ⇒(2/(cos^2 y))=(1/(cos^2 x))  ⇒2cos^2 x=cos^2 y  ⇒cos 2x+sin^2 y=0
$$\mathrm{tan}\:^{\mathrm{2}} {x}=\mathrm{1}+\mathrm{2tan}\:^{\mathrm{2}} {y} \\ $$$$\Rightarrow\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}=\mathrm{2sec}\:^{\mathrm{2}} {y}−\mathrm{1}\:\:\:\:\: \\ $$$$\Rightarrow\frac{\mathrm{2}}{\mathrm{cos}\:^{\mathrm{2}} {y}}=\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {x}} \\ $$$$\Rightarrow\mathrm{2cos}\:^{\mathrm{2}} {x}=\mathrm{cos}\:^{\mathrm{2}} {y} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2}{x}+\mathrm{sin}\:^{\mathrm{2}} {y}=\mathrm{0} \\ $$
Answered by Ankushkumarparcha last updated on 12/Apr/21
Solution: tan^2 (x) = 1+2tan^2 (y)  (given)  tan^2 (x) = ((1+sin^2 (y))/(cos^2 (y))) => 2sec^2 (y)−1  sec^2 (x) = 2sec^2 (y) => −sin^2 (y) = cos(2x)  cos(2x)+sin^2 (y) = 0
$${Solution}:\:\mathrm{tan}^{\mathrm{2}} \left({x}\right)\:=\:\mathrm{1}+\mathrm{2tan}^{\mathrm{2}} \left({y}\right)\:\:\left({given}\right) \\ $$$$\mathrm{tan}^{\mathrm{2}} \left({x}\right)\:=\:\frac{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \left({y}\right)}{\mathrm{cos}^{\mathrm{2}} \left({y}\right)}\:=>\:\mathrm{2sec}^{\mathrm{2}} \left({y}\right)−\mathrm{1} \\ $$$$\mathrm{sec}^{\mathrm{2}} \left({x}\right)\:=\:\mathrm{2sec}^{\mathrm{2}} \left({y}\right)\:=>\:−\mathrm{sin}^{\mathrm{2}} \left({y}\right)\:=\:\mathrm{cos}\left(\mathrm{2}{x}\right) \\ $$$$\mathrm{cos}\left(\mathrm{2}{x}\right)+\mathrm{sin}^{\mathrm{2}} \left({y}\right)\:=\:\mathrm{0} \\ $$

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