Evaluate each of the following: `sec^(-1)(sec(13pi)/4)` (ii) `sec^(-1)(sec(25pi)/6)`

Explanation

(i) We know that

`\sec ^{-1}(\sec \theta)=\theta, \quad[0, \pi / 2) \cup(\pi / 2, \pi]`

We have


`\sec ^{-1}(\sec \frac{13 \pi}{4})=\sec ^{-1}[\sec (4 \pi-\frac{3 \pi}{4})]`

`=\sec ^{-1}[\sec (\frac{3 \pi}{4})]`

`=\frac{3 \pi}{4}`




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