Let
x=(1+tan1)(1+tan2)....(1+tan25)
y=(1-tan136)(1-tan137)....(1-tan160)
Then xy equals to
Answers
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
Learn more:
tan⁻¹ 15 +tan⁻¹ 17 +tan⁻¹ 13 + tan⁻¹ 18 = π4 सिद्ध कीजिए
https://brainly.in/question/16556304
Prove that tan inverse x +tan inverse y=tan inverse x+y by 1-xy
https://brainly.in/question/6632911
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