How do you prove $tan(2A)/(1+sec(2A))=tan(A)$ ?

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Answer 1 · 2015-07-06T11:53:39

My proof is done from RHS to LHS.

$tanA=sinA/cosA$

multiply and divide by 2cosA

$tanA=(2sinAcosA)/(2cos^2A)$

$tanA=(Sin2A)/(cos^2A+cos^2A)$

$tanA=(sin2A)/(1-sin^2A+cos^2A)$

$tanA=(sin2A)/(1+cos2A)$

Divide numerator and denominator with cos2A

$tanA=((sin2A)//cos2A)/((1+cos2A)//cos2A$

$tanA=(tan2A)/(1+sec2A)$

Hence proved

Answer 2 · 2015-07-06T16:55:27

Prove trig identity

$(tan 2a)/(1 + (1/cos (2a)) =$ $((sin 2a)/(cos 2a)).((cos 2a)/(cos 2a + 1))$

= $(sin 2a)/(cos 2a + 1)$ = $(2sin acos a)/(2cos^2 a) =$

$= sin a/cos a = tan a$

How do you prove tan(2A)/(1+sec(2A))=tan(A)tan(2A)/(1+sec(2A))=tan(A)?