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if secA-tanA=?3-2. show that (1-sinA)/(1+sinA)=7-4?3

LHS:
seca-tanA=?3-2           (?- refers to root)
1-sinA/1+sinA =LHS
divide the numerator and denominator by cosA
​you will get
secA-tanA/secA+tanA = LHS
​then square on both sides,
you will get
(secA-tanA)^2/1           (using 1+tan^2A=sec^2A)
therefore,
(?3-2)^2
=7-4?3 (expand the value above)


 
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View Full Answer
 Given :  ​​secA - tanA = √3 -2 
          = 1/cosA - sinA/cosA = √3 -2
          =  (1-sinA)/cosA = √3 -2
          =   [(1-sinA/cosA)]2  = (√3 -2)2     (squaring both sides)
          =  (1-sinA)2/ cos2A  = (3 + 4 - 4√3)
          =   (1-sinA)2/ 1-sin2A  = 7 - 4√3
          =   (1-sinA)(1-sinA)/  (1-sinA)(1+sinA) =  7 - 4√3
         =   (1-sinA)/(1+sinA)  =  7 - 4√3
HENCE PROVED
Hope it helped!!
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