If (1+\tan 1^{\circ})(1+\tan 2^{\circ}) . . .(1+\tan 45^{\circ})=2^{n}, then the value of n is

NIMCET 2008 Question Paper with solutions

Question : 27 of 120

Marks: +1, -0

(1+\tan 1^{∘})(1+\tan 2^{∘}) . . .(1+\tan 45^{∘})=2^{n}

then the value of

n

Solution:

As,

\tan 45^{∘}=\tan [x+(45^{∘}-x)]

=\;{\tan x+\tan (45^{∘}-x)}/{1-\tan x \tan (45^{∘}-x)}=1

⇒ 1-\tan x \tan (45^{∘}-x)=\tan x+\tan (45^{∘}-x)

⇒ 1=\tan x+\tan (45^{∘}-x)+\tan x \tan (45^{∘}-x)

⇒ 2=(1+\tan x)(1+\tan (45^{∘}-x))

....(i)
So.

(1+\tan 1^{∘})(1+\tan 44^{∘})=2

(1+\tan 2^{∘})(1+\tan 43^{9})=2

(1+\tan 22^{∘})(1+\tan 23^{9})=2

and

(1+\tan 45^{∘})=2

⇒[(1+\tan 1^{∘})(1+\tan 2^{9}) . . . .22 {times})(1+\tan 45^{∘})

=(2 ⋅ 2 ⋅ 2 . . . .22

times)

⋅ 2=z^{22} ⋅ 2=2^{23}

⇒ n=23

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