Question

If(1+tan5)(1+tan10).....(1+tan45)=2^k

Then k=?

Answer 1

Answer:

we know :- tan45° = 1, and tan(A + 45°) =

then, tan(A + 45)(1 - tanAtan45) = (tanA + 1)

Now, (1 + tan5)(1 + tan10)(1 + tan15).........(1 + tan45) = 2^k

(1 + tan5)(1 + tan40) * (1 + tan10)(1 + tan35) * (1 + tan15)(1 + tan30) * (1 + tan20)(1 + tan25) * (1 + 1) = 2^k

[1 + tan40tan5 + (tan5 + tan40)] * [1 + tan10tan35 + (tan35 + tan10)] * [1 + tan20tan25 + (tan20 + tan25)] * 2 = 2^k

[1 + tan40tan5 + tan(40 + 5)(1 - tan40tan5)] * [1 + tan10tan35 + tan(10 + 35)(1 - tan10tan35)] * [1 + tan20tan25 + tan(20 + 25)(1 - tan20tan25)] * 2 = 2^k

[1 + tan40tan5 + tan45(1 - tan40tan5)] * [1 + tan10tan35 + tan45(1 - tan10tan35)] * [1 + tan20tan25 + tan45(1 - tan20tan25)] * 2 = 2^k

[1 + tan40tan5 + 1 - tan40tan5] * [1 + tan10tan35 + 1 - tan10tan35] * [1 + tan20tan25 + 1 - tan20tan25] * 2 = 2^k

2 * 2 * 2 * 2 = 2^k

2^4 = 2^k

On comparing both side of base,

we get , k = 4

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