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

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

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Answer 1 · 2015-07-06T11:53:39

My proof is done from RHS to LHS.

Explanation:

$tan A = \frac{sin A}{cos A}$ $tanA=sinA/cosA$

multiply and divide by 2cosA

$tan A = \frac{2 sin A cos A}{2 {cos}^{2} A}$ $tanA=(2sinAcosA)/(2cos^2A)$

$tan A = \frac{sin 2 A}{{cos}^{2} A + {cos}^{2} A}$ $tanA=(Sin2A)/(cos^2A+cos^2A)$

$tan A = \frac{sin 2 A}{1 - {sin}^{2} A + {cos}^{2} A}$ $tanA=(sin2A)/(1-sin^2A+cos^2A)$

$tan A = \frac{sin 2 A}{1 + cos 2 A}$ $tanA=(sin2A)/(1+cos2A)$

Divide numerator and denominator with cos2A

$tan A = \frac{(sin 2 A) / cos 2 A}{(1 + cos 2 A) / cos 2 A}$ $tanA=((sin2A)//cos2A)/((1+cos2A)//cos2A$

$tan A = \frac{tan 2 A}{1 + sec 2 A}$ $tanA=(tan2A)/(1+sec2A)$

Hence proved

Answer 2 · 2015-07-06T16:55:27

Prove trig identity

Explanation:

$\frac{tan 2 a}{1 + (\frac{1}{cos (2 a)}) =}$ $(tan 2a)/(1 + (1/cos (2a)) =$ $(\frac{sin 2 a}{cos 2 a}) . (\frac{cos 2 a}{cos 2 a + 1})$ $((sin 2a)/(cos 2a)).((cos 2a)/(cos 2a + 1))$

= $\frac{sin 2 a}{cos 2 a + 1}$ $(sin 2a)/(cos 2a + 1)$ = $\frac{2 sin a cos a}{2 {cos}^{2} a} =$ $(2sin acos a)/(2cos^2 a) =$

$= \frac{sin a}{cos a} = tan a$ $= sin a/cos a = tan a$

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

2 Answers

Explanation:

Explanation:

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