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