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