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Question Number 39231 by LX Z last updated on 04/Jul/18
(sin^(−1) x)^2 +  (sin^(−1) y)^2 +  2(sin^(−1) x)  (sin^(−1) y)= π^2 ,then x^2 +y^2  is equal to?
$$\left({sin}^{−\mathrm{1}} {x}\right)^{\mathrm{2}} +\:\:\left({sin}^{−\mathrm{1}} {y}\right)^{\mathrm{2}} +\:\:\mathrm{2}\left({sin}^{−\mathrm{1}} {x}\right) \\ $$$$\left({sin}^{−\mathrm{1}} {y}\right)=\:\pi^{\mathrm{2}} ,{then}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:{is}\:{equal}\:{to}? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jul/18
a=sin^(−1) x    b=sin^(−1) y       a^2 +2ab+b^2 =Π^2   a+b=Π  sin^(−1) x+sin^(−1) y=Π  sin^(−1) x=Π−sin^(−1) y  a+b=Π  sina=sin(Π−b)  sina=sinb  a=b     a+b=Π   a=(Π/2)     b=(Π/2)  sin^(−1) x=(Π/2)    x=1  sin^(−1) y=(Π/2)    y=1  x^2 +y^2 =2  pls check
$${a}={sin}^{−\mathrm{1}} {x}\:\:\:\:{b}={sin}^{−\mathrm{1}} {y}\:\:\:\:\: \\ $$$${a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} =\prod^{\mathrm{2}} \\ $$$${a}+{b}=\Pi \\ $$$${sin}^{−\mathrm{1}} {x}+{sin}^{−\mathrm{1}} {y}=\Pi \\ $$$${sin}^{−\mathrm{1}} {x}=\Pi−{sin}^{−\mathrm{1}} {y} \\ $$$${a}+{b}=\Pi \\ $$$${sina}={sin}\left(\Pi−{b}\right) \\ $$$${sina}={sinb} \\ $$$${a}={b}\:\:\: \\ $$$${a}+{b}=\Pi\:\:\:{a}=\frac{\Pi}{\mathrm{2}}\:\:\:\:\:{b}=\frac{\Pi}{\mathrm{2}} \\ $$$${sin}^{−\mathrm{1}} {x}=\frac{\Pi}{\mathrm{2}}\:\:\:\:{x}=\mathrm{1} \\ $$$${sin}^{−\mathrm{1}} {y}=\frac{\Pi}{\mathrm{2}}\:\:\:\:{y}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$${pls}\:{check}\: \\ $$$$ \\ $$

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