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प्रश्न
If `(sin^-1x)^2+(cos^-1x)^2=(17pi^2)/36,` Find x
उत्तर
`(sin^-1x)^2+(cos^-1x)^2=(17pi^2)/36`
⇒ `(sin^-1x)^2+(pi/2-sin^-1x)^2=(17pi^2)/36`
Let `sin^-1x=y`
`therefore(y)^2+(pi/2-y)^2=(17pi^2)/36`
⇒ `y^2+pi^2/4+y^2-2xxpi/2xxy=(17pi^2)/36`
⇒ `2y^2-piy=(2pi^2)/9`
⇒ `18y^2-9piy-2pi^2=0`
⇒ `18y^2-12piy+3piy-2pi^2=0`
⇒ `6y(3y-2pi)+pi(3y+2pi)=0`
⇒ `(3y-2pi)(6y+pi)=0`
⇒ `y=pi/6` [Neglecting `y=2/3pi` as it is not satisfying the question]
`thereforex=siny=sin(-pi/6)=-1/2`
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