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प्रश्न
Find the principal value of the following:
`sec^-1(-sqrt2)`
उत्तर
Let `sec^-1(-sqrt2)=y`
Then,
`secy=-sqrt2`
We know that the range of the principal value branch is [0, π] - `{pi/2}`.
Thus,
`secy=-sqrt2=sec((3pi)/4)`
`=>y=(3pi)/4in[0,pi],y!=pi/2`
Hence, the principal value of `sec^-1(-sqrt2) is (3pi)/4.`
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