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Question
If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,` then write the value of x + y + z.
Solution
`sin^-1x+sin^-1y+sin^-1z=(3pi)/2`
`=>sin^-1x+sin^-1y+sin^-1z=pi/2+pi/2+pi/2` [As the maximum value in the range of `sin^-1x is pi/2` And here sum of three inverse of sine is 3 times `pi/2`. i.e., every sin inverse function is equal to `pi/2` here.]
`=>sin^-1x=pi/2,sin^-1y=pi/2andsin^-1z=pi/2`
`=>x=1,y=1andx=1`
`thereforex+y+z=1+1+1=3`
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