Math, asked by AravindGajula, 6 months ago

The value of cot(π÷24) is
1) 1-√2 + √3+ √6
2)2+√2+√3-√6
3) 1+√2+√3+√6
4)2+√2+√3+√6

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Answered by mutasimsheikh28947

Answer:

4eo7fbz4eo74eo4e4eo7fbzgnxd4eo7fbzgnxdtjesztentes4eo7fbzg4eoirursYRR06RYYISEsYRR06RYYISEI5FESIR6ITUL7YDNYJYTURTEYKUKRKUIR6I7fbz4eo74eo4e4eo7fbzgnxd4eo7fbzgnxdtjesztentes4eo7fbzg4eoirursYRR06RYYISEI5FMHTKYRR06RYYISEI5FESIR6ITUL7YDNYJYTURTEYKUKRKUIR6I7fbz4eo74eo4e4eo7fbzgnxd4eo7fbzgnxdtjesztentes4eo7fbzg4eoirursYRR06RYYISEI5FMHTK5RR06RYYISEI5FESIR6ITUL7YDNYJYTURTEYKUKRKUIR6I7fbz4eo74eo4e4eo7fbzgnxd4eo7fbzgnxdtjesztentes4eo7fbzg4eoirursYRR06RYYISEI5FMHTK5ITIKRIMRIURIURUNYRYRHYRNTH47K7TIMTUMTHYRJ473YRTURYIHKOFTTIIPT7976IRER795R7OR68E6I635358E68E4E694E68UIR6I

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The value of cot(π÷24) is1) 1-√2 + √3+ √62)2+√2+√3-√63) 1+√2+√3+√64)2+√2+√3+√6​

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The value of cot(π÷24) is
1) 1-√2 + √3+ √6
2)2+√2+√3-√6
3) 1+√2+√3+√6
4)2+√2+√3+√6