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प्रश्न
If `sin^-1x+sin^-1y=pi/3` and `cos^-1x-cos^-1y=pi/6`, find the values of x and y.
उत्तर
`cos^-1x-cos^-1y=pi/6`
⇒ `pi/2-sin^-1x-pi/2+sin^-1y=pi/6` `[thereforecos^-1x=pi/2-sin^-1x]`
⇒ `-(sin^-1x-sin^-1y)=pi/6`
⇒ `sin^-1x-sin^-1y=-pi/6`
Solving `sin^-1x+sin^-1y=pi/3` and `sin^-1x-sin^-1y=-pi/6` we will get `2sin^-1x=pi/6`
⇒ `sin^-1x=pi/12`
⇒ `x=sin pi/12=(sqrt3-1)/(2sqrt2)`
and
`sin^-1y=pi/3-sin^-1x`
⇒ `sin^-1y=pi/3-pi/12`
⇒ `sin^-1y=pi/4`
⇒ `y=sin pi/4=1/sqrt2`
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