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If x=SecA-TanA y=CosecA+CotA P.t x.y+1=y-x

Asked by chethan U 22nd July 2012, 5:37 PM
Answered by Expert
Answer:
Answer : Given : If x=SecA-TanA y=CosecA+CotA
To Prove : x.y + 1 = y - x
 
LHS 
= xy + 1
= ((sec A - tan A ) (cosec A + cotA) )+ 1
= sec A cosec A + sec A cotA - tanA cosecA - tan A cotA + 1 
 
{  using cotA = 1/ tan A  in 4th term
, tanA =sinA / cosA and cosec A = 1/sinA in 3rd term,
  sec A = 1/cosA and cotA = cosA / sinA in 2nd term }
 
=secA cosecA + ( (1/cosA) ( cos A / sin A ) ) - ( (1/sinA )(sinA / cos A ) )   - tan A (1/tanA) + 1
 
= secA cosecA +(1/sinA ) -(1/cosA ) - 1 +1
= secA cosecA +(1/sinA ) -(1/cosA ) 
 
{ using 1/sinA = cosec A and 1/cos A = sec A , we get}
 
= secA cosecA + cosecA - secA .......................(1) 
 
RHS 
= y- x
= ( cosec A + cotA ) - ( sec A - tanA) 
= cosecA - sec A  + ( cotA + tanA ) 
 
{ using cotA = 1/tanA }
 
= cosecA -secA +( (1/tanA) + tanA )
= cosecA -secA + ( (1+tan2A) /tanA )
using 1+tan2A = sec2
 
=cosecA -secA + ( sec2A /tanA) 
 using secA = 1/cos A and  tan A = sinA / cosA  , we get 
 
= cosecA -secA + ( ( 1/cos2 A) * (cosA/sinA) )
= cosecA - secA + (  1/ (cos A sinA ) )
= cosecA - secA + cosecA secA..............(2)   { using 1/sinA = cosecA and 1/cosA=secA }
 
from eq (1) and (2) , LHS =RHS 
Hence Proved
Answered by Expert 22nd July 2012, 6:53 PM
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