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Given :- (1) TanA + SinA = m
(2) TanA - SinA = n
To prove :- m2 - n2 = 4 x (root mn)
Proof :-
L.H.S = m2 - n2
= (TanA + SinA)2 - (TanA - SinA)2
= (Tan2A + Sin2A + 2 TanA SinA) - ( Tan2A + Sin2A - 2 TanA SinA)
= Tan2A + Sin2A + 2 TanA SinA - Tan2A - Sin2A + 2 TanA SinA
= Tan2A + Sin2A + 2 TanA SinA - Tan2A - Sin2A + 2 TanA SinA
= 4 TanA SinA
R..H.S ;- 4 root mn
= 4 x [ root (TanA + SinA)(TanA - SinA) ]
= 4 x [ root ( Tan2A - Sin2A ) ]
= 4 x [ root ( Sin2A / Cos2A - Sin2A )
= 4 x [ root { (Sin2A - Sin2ACos2A) / Cos2A } ]
= 4 x [ root { Sin2A (1 - Cos2A) } / Cos2A ]
= 4 x [ root { (Sin2A x Sin2A ) / Cos2A ]
= 4 x (root [ Sin2A Tan2A ] )
= 4 TanA SinA = L.H.S
Hence Proved
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