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11.The value of log cotl+ log cot2++ log cot 89, İs-a) tan I, b) 0, c), d) tan89tr

Answer»

log(cot1) +log(cot2) +....log(cot89)= log[cot1×cot2×cot3×.....cot88×cot89]= log[cot1×cot2×.....×tan(90-88)×tan(90-89)]= log[cot1×cot2×.....×tan(2)×tan(1)]

but cotx.tanx = 1 , so here all the cot and tan will multiply to get 1 leaving the middle term cot45 = 1

so, log[cot (45)]=> log[1] = 0



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