If cos θ =
, verify that tan2θ – sin2θ = tan2θ⋅ sin2θ

Let ∠B = θ
and BC = 2√2 and AB = 3
⇒ AC = 1 (by Pythagoras theorem)
we know that
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⇒ tanθ =![]()
⇒ sinθ =![]()
L.H.S
tan2
– sin2![]()
by the above values of tanθ and sinθ
tan2
– sin2![]()

![]()
R.H.S
![]()
by the above values of tanθ and sinθ
tan2
× sin2![]()

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prove that 3cosB – 4cos3B = 0.