Question

log (cos((x-a) (x-b)(x-c) ............... (x-z)) ​

Answer 1

log[cos{(x - a) (x - b) ... (x - z)}] = 0

Step-by-step explanation:

Let, P = (x - a) (x - b) (x - c) ... (x - z)

Since a, b, c, ..., z are the letters from the English alphabet, there is a term (x - x) in P which can be written as follows

P = (x - a) (x - b) (x - c) ... (x - x) (x - y) (x - z)

= 0, since (x - x) = 0

i.e., P = 0

Taking cosine in both sides, we get

cosP = cos0

or, cosP = 1, since cos0 = 1

Now taking logarithm in both sides, we get

log(cosP) = log1

or, log[cos{(x - a) (x - b) ... (x - z)}] = 0

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Answer 2

Answer:

step 1:log[cos(x-a)(x-b)..........(x-z)]

step 2:take" x-" as common from (x-a)(x-b).....(x-z)

step 3:After taking "x-" as common then we get

log[cos(x-(a.b.c......z)]

step 4: Consider (a.b.c.....z) as "x"

step 5:log[cos( x-x)]

step 6:log[cos(0)]

step 7:log1

step 8:0