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प्रश्न
For the principal value, evaluate of the following:
`sin^-1(-sqrt3/2)+cos^-1(sqrt3/2)`
उत्तर
`sin^-1(-sqrt3/2)+cos^-1(sqrt3/2)`
`=sin^-1{sin(-pi/2)}+cos^-1(cos pi/6)`
`=-pi/3+pi/6` `[because"Range of sine is"[-pi/2,pi/2]; -pi/3 in [-pi/2, pi/2] "and range of cosine is" [0,pi] ; pi/6 in [0, pi]]`
`=-pi/6`
`therefore sin^-1(-sqrt3/2)+cos^-1(sqrt3/2)=-pi/6`
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