Question

(1 + tan1)(1+tan2)(1+tan3).................(1+tan45) = 2^n
then find the value of n

a) 22 b)24 c)23 d)12
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Answer 1

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Answer 2

Solution :-

We have got the series :-

(1 + tan1°)(1 + tan2°)(1 + tan3°) ....

(1 + 45°) = 2ⁿ

Now let us start solving LHS :-

$= \small{(1 + tan1^{\circ})(1 + tan2^{\circ})(1 + tan3^{\circ}) ....(1 + 45^{\circ})}$

It can be grouped together like

$= \small{\left[\left((1 + tan1^{\circ})(1 + tan44^{\circ})\right)\left((1 + tan2^{\circ})(1+ tan43^{\circ})\right) ....\right](1 + 45^{\circ})}$

Now we may go through this :-

If A + B = 45° then

$( 1 + tanA )(1 + tanB)$

$= ( 1 + tanA )(1 + tan(45^{\circ} - B))$

$= ( 1 + tanA) \left( 1 + \dfrac{ tan45^{\circ} - tanA}{ 1 + tan45^{\circ}tanA}\right)$

$= ( 1 + tanA) \left( 1 + \dfrac{ 1 - tanA}{ 1+ tanA}\right)$

$= ( 1 + tanA) \left( \dfrac{1 + tanA 1 - tanA}{ 1+ tanA}\right)$

$= ( 1 + tanA) \left( \dfrac{ 2}{ 1+ tanA}\right)$

$= 2$

So the product of two Exept the one with 45° will give 2.

Therefore for there will be

→ 44 ÷ 2

→ 22 times 2

$= 2^{22} \times ( 1 tan45^{\circ})$

$= 2^{22} \times ( 1 + 1 )$

$= 2^{22} \times 2$

$= 2^{23}$

Hence

n = 23