Question

Let
x = (1 + tan 1°)(1+tan 2°)... (1 + tan 25
and
y=(1 - tan 136°)(1 – tan 137°)... (1 - ta
then the value of xy equals​

Answer 1

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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