Question

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

Answer 1

Explanation:

=(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

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 + tan1)(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)

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

=> xy  = 2²⁵

xy  = 2²⁵