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Question Number 24973 by Rishabh#1 last updated on 30/Nov/17
If a sin^2 x+b cos^2 x=c, b sin^2 y+a cos^2 y=d and a tan x= b tan y then (a^2 /b^2 ) =? (in terms of a,b,c,d)
$$\mathrm{If}\:{a}\:\mathrm{sin}^{\mathrm{2}} {x}+{b}\:\mathrm{cos}^{\mathrm{2}} {x}={c},\:{b}\:\mathrm{sin}^{\mathrm{2}} {y}+{a}\:\mathrm{cos}^{\mathrm{2}} {y}={d} \\ $$$$\mathrm{and}\:\:{a}\:\mathrm{tan}\:{x}=\:{b}\:\mathrm{tan}\:{y}\:\mathrm{then}\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=? \\ $$$$\left(\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{a},{b},{c},{d}\right) \\ $$
Commented by nnnavendu last updated on 30/Nov/17
ans your question is incomplete please resend question bsin^2 y+acos^2 y=?
$$ \\ $$$$ \\ $$$$\mathrm{ans} \\ $$$$\mathrm{your}\:\mathrm{question}\:\mathrm{is}\:\mathrm{incomplete} \\ $$$$\mathrm{please}\:\:\mathrm{resend}\:\mathrm{question} \\ $$$$\mathrm{bsin}^{\mathrm{2}} \mathrm{y}+\mathrm{acos}^{\mathrm{2}} \mathrm{y}=? \\ $$
Commented by Rishabh#1 last updated on 30/Nov/17
I have updated the question. Thank you
$$\mathrm{I}\:\mathrm{have}\:\mathrm{updated}\:\mathrm{the}\:\mathrm{question}.\:\mathrm{Thank}\:\mathrm{you} \\ $$
Answered by nnnavendu last updated on 30/Nov/17
ans asin^2 x+bcos^2 x=c a(1−cos^2 x)+bcos^2 x=c a−acos^2 x+bbcos^2 x=c (a−c)=cos^2 x(a−b) cos^2 x=((a−c)/(a−b)) sin^2 x=1−cos^2 x =1−(((a−c)/(a−b))) =((a−b−a+c)/(a−b)) =((c−b)/(a−b)) now bsin^2 y+acos^2 y=d b(1−cos^2 y)+acos^2 y=d b−bcos^2 y+acos^2 y=d (b−d)=cos^2 y(b−a) cos^2 y=((b−d)/(b−a)) sin^2 y=1−cos^2 y =1−(((b−d)/(b−a))) =((b−a−b+d)/(b−a)) =((d−a)/(b−a)) now a tan x= b tan y ((a/b))^2 =(((tany)/(tanx)))^2 (a^2 /b^2 )=((sin^2 y)/(cos^2 y))×((cos^2 x)/(sin^2 x)) =((((d−a)/(b−a))/((b−d)/(b−a))))×((((a−c)/(a−b))/((c−b)/(a−b)))) =(((d−a)/(b−d)))×(((a−c)/(c−b))) by natural navendu
$$\mathrm{ans} \\ $$$$ \\ $$$$\mathrm{asin}^{\mathrm{2}} \mathrm{x}+\mathrm{bcos}^{\mathrm{2}} \mathrm{x}=\mathrm{c} \\ $$$$\mathrm{a}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)+\mathrm{bcos}^{\mathrm{2}} \mathrm{x}=\mathrm{c} \\ $$$$\mathrm{a}−\mathrm{acos}^{\mathrm{2}} \mathrm{x}+\mathrm{bbcos}^{\mathrm{2}} \mathrm{x}=\mathrm{c} \\ $$$$\left(\mathrm{a}−\mathrm{c}\right)=\mathrm{cos}^{\mathrm{2}} \mathrm{x}\left(\mathrm{a}−\mathrm{b}\right) \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{x}=\frac{\mathrm{a}−\mathrm{c}}{\mathrm{a}−\mathrm{b}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{x}=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\left(\frac{\mathrm{a}−\mathrm{c}}{\mathrm{a}−\mathrm{b}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{a}−\mathrm{b}−\mathrm{a}+\mathrm{c}}{\mathrm{a}−\mathrm{b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{c}−\mathrm{b}}{\mathrm{a}−\mathrm{b}} \\ $$$$\:\mathrm{now} \\ $$$$\mathrm{bsin}^{\mathrm{2}} \mathrm{y}+\mathrm{acos}^{\mathrm{2}} \mathrm{y}=\mathrm{d} \\ $$$$\mathrm{b}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{y}\right)+\mathrm{acos}^{\mathrm{2}} \mathrm{y}=\mathrm{d} \\ $$$$\mathrm{b}−\mathrm{bcos}^{\mathrm{2}} \mathrm{y}+\mathrm{acos}^{\mathrm{2}} \mathrm{y}=\mathrm{d} \\ $$$$\left(\mathrm{b}−\mathrm{d}\right)=\mathrm{cos}^{\mathrm{2}} \mathrm{y}\left(\mathrm{b}−\mathrm{a}\right) \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{y}=\frac{\mathrm{b}−\mathrm{d}}{\mathrm{b}−\mathrm{a}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{y}=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\left(\frac{\mathrm{b}−\mathrm{d}}{\mathrm{b}−\mathrm{a}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{b}−\mathrm{a}−\mathrm{b}+\mathrm{d}}{\mathrm{b}−\mathrm{a}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{d}−\mathrm{a}}{\mathrm{b}−\mathrm{a}} \\ $$$$ \\ $$$$\mathrm{now} \\ $$$${a}\:\mathrm{tan}\:{x}=\:{b}\:\mathrm{tan}\:{y}\: \\ $$$$\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{tany}}{\mathrm{tanx}}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }=\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{y}}{\mathrm{cos}^{\mathrm{2}} \mathrm{y}}×\frac{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}} \\ $$$$\:\:\:\:\:\:=\left(\frac{\frac{\mathrm{d}−\mathrm{a}}{\mathrm{b}−\mathrm{a}}}{\frac{\mathrm{b}−\mathrm{d}}{\mathrm{b}−\mathrm{a}}}\right)×\left(\frac{\frac{\mathrm{a}−\mathrm{c}}{\mathrm{a}−\mathrm{b}}}{\frac{\mathrm{c}−\mathrm{b}}{\mathrm{a}−\mathrm{b}}}\right) \\ $$$$\:\:\:\:\:\:=\left(\frac{\mathrm{d}−\mathrm{a}}{\mathrm{b}−\mathrm{d}}\right)×\left(\frac{\mathrm{a}−\mathrm{c}}{\mathrm{c}−\mathrm{b}}\right) \\ $$$$\:\:\:\:\mathrm{by}\:\mathrm{natural}\:\mathrm{navendu} \\ $$
Commented by Rishabh#1 last updated on 30/Nov/17
thanks a lot.
$${thanks}\:{a}\:{lot}. \\ $$

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