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If 3 cot A = 4, check whether (1-tan2 A)/(1+tan2 A) = cos2 A - sin2 A or not

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if  3cotA= 4 check whether
1-tan2A/1+tan2A =  cos2A – sin2A or not.
Ans:
 [note:You can solve it without using value]
RHS = cos2A – sin2A
[using  formula sin2A + cos2A = 1 hence cos2A =  1- sin2 ]
  ∴RHS = 1-sin2A – sin2A
            = 1- 2sin2A ………… (i)
LHS = 1-tan2A/1+tan2A
      = [1– sin2A/cos2A] /[1+ sin2A/cos2A]
      = [cos2A – sin2A/cos2A] /  
                                [cos2A+sin2A/cos2A ]
     = cos2A – sin2A/cos2A X
                                  cos2A /cos2A+sin2A
   = cos2A – sin2A / cos2A+sin2A
[using  formula sin2A + cos2A = 1]
  = 1-sin2A –sin2A / 1- sin2A +sin2A
  = 1- 2 sin2A / 1
 = 1- 2 sin2A  …………….. (ii)
From Eqn (i) & (ii) it is proved that
LHS = RHS proved and hence
1-tan2A/1+tan2A =  cos2A – sin2A

1 Answer

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Best answer

Solution:  Given
Cot A=4/3 Or B/P=4/3
Let B=4k and P=3k
So in a right angle triangle with angle A
P2 + B2 =H2
H=5k
Now tan A=1/cot A=3/4
Cos A=B/H=4/5
Sin A=P/H=3/5
Let us take the LHS

RHS = cos2 A -  sin2A= 7/25
So LHS=RHS,so the statement is true

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