Question

Let
x=(1+tan1)(1+tan2)....(1+tan25)
y=(1-tan136)(1-tan137)....(1-tan160)
Then xy equals to

Answer 1

=(1+tan 0o)(1 + tan1°) (1 + tan2°) (1 + tan3°) ………. (1 + tan 45°) {Since , tan 0o = 0}

Now , we know that

tan(A + B) = [ tan(A) + tan(B) ] / [ 1 - tan(A) tan(B) ]

And tan(45) = 1

So

tan(45) = [ tan1 + tan44 ] / [ 1 - tan1 tan44 ]

tan(45) = [ tan2 + tan43 ] / [ 1 - tan2 tan43 ]

.

.

.

tan(45) = [ tan22 + tan23 ] / [ 1 - tan22 tan23 ]

So we have ,

tan1 + tan44 = tan(45) [ 1 - tan1 tan44 ]

tan2 + tan43 = tan(45) [ 1 - tan2 tan43 ]

.

.

.

tan22 + tan23 = tan(45) [ 1 - tan22 tan23 ]

So finally we have ,

(1 + tan1) (1 + tan2) (1 + tan3) ... (1 + tan44) (1 + tan45)

= (1 + tan1) (1 + tan2) (1 + tan3) ... (1 + tan44) (1 + 1)

= 2 (1 + tan1) (1 + tan2) (1 + tan3) ... (1 + tan44)

= 2 (1 + tan1) (1 + tan44) * (1 + tan2) (1 + tan43) * (1 + tan3) (1 + tan42) * ... * (1 + tan22) (1 + tan23)

= 2 (tan1 + tan44 + tan1 tan44 + 1) (tan2 + tan43 + tan2 tan43 + 1) ... (tan22 + tan23 + tan22 tan23 + 1)

= 2 ((tan45)(1 - tan1 tan44) + tan1 tan44 + 1) ((tan45)(1 - tan2 tan43) + tan2 tan43 + 1) ... ((tan45)(1 - tan22 tan23) + tan22 tan23 + 1)

= 2 (1 - tan1 tan44 + tan1 tan44 + 1) (1 - tan2 tan43 + tan2 tan43 + 1) ... (1 - tan22 tan23 + tan22 tan23 + 1)

= 2 (2) (2) ... (2)

= 2 (2^22)

= 2^23

Answer 2

Let

x=(1+tan1)(1+tan2)....(1+tan25)

y=(1-tan136)(1-tan137)....(1-tan160)

Then xy equals to

Given : x=(1+tan1°)(1+tan2°)....(1+tan25°) , y=(1-tan136°)(1-tan137°)....(1-tan160°)

To find : xy

Solution:

x = (1+tan1°)(1+tan2°)........(1+tan25°)

y = (1-tan136°)(1-tan137°)....(1-tan160°)

tan 136° = -tan44° , tan137° = -tan43° ,.......tan160° = - tan20°

=> y = ( 1 + tan44°)(1 + tan43°)...................(1 + tan20°)

xy = ((1+tan1°)(1+tan2°)........(1+tan25°))(( 1 + tan44°)(1 + tan43°)...................(1 + tan20°))

=> xy = (1 + tan1v)(1 + Tan44°)(1 + tan2°)(1 + tan43°) ....................(1 + tan25°)((1 + tan20°)

now using concept

Tan ( A + B) = (TanA + TanB )/(1 - TanATanB) such that A + B = 45°

=> 1 = (TanA + TanB )/(1 - TanATanB)

=> 1 - TanATanB = TanA + TanB

=> 1 = TanA + TanB + TanATanB

=> 1 + 1 = 1 + TanA + TanB + TanATanB

=> 2 = ( 1 + TanA) + TanB(1 + TanA)

=> 2 = (1 + TanA)(1 + TanB)

where A + B = 45°

=> xy = (2)(2) ....................................(2) ( 25 times)

=> xy = 2²⁵

xy = 2²⁵

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