If $(1 + tan 1^{o}) (1 + tan 2^{o}) (1 + tan 3^{o}) . . . . (1 + tan 45^{o}) = 2^{n}$ , then $n$ is equal to

21

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24

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23

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22

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Solution

The correct option is C $23$
As we know that

$(1 + t a n x) (1 + t a n (45 - x)) = (1 + t a n x) \frac{(1 + t a n 45 t a n x + t a n 45 - t a n x)}{1 + t a n x} = 2$

So $(1 + t a n 1) (1 + t a n 44) (1 + t a n 2) (1 + t a n 43) + . . . . . . . (1 + t a n 45) = 2 \times \times . . . . . . \times 2 \times (1 + 1)$ ( using above formula)

= $2^{22} \times 2 = 2^{23}$

so we get n=23

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Similar questions

(1 + tan 1^{o}) (1 + tan 2^{o}) (1 + tan 3^{o}) . . . . . . . . . . . . (1 + tan 45^{o})

(1 + tan 1^{o}) (1 + tan 2^{o}) . . . (1 + tan 45^{o}) = 2^{25 - n}

n

1 \cdot 2^{1} + 2 \cdot 2^{2} + 3 \cdot 2^{3} + 4 \cdot 2^{4} + \dots + n \cdot 2^{n} = 2 + 2^{n + 5},

(2 \times 2^{2}) + (3 \times 2^{3}) + (4 \times 2^{4}) . . . + (n \times 2^{n}) = 2^{n + 10}

n =

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