Respuesta :

[tex] \cos(2A)=\cos^2 A-\sin^2A=(\cos A+ \sin A)(\cos A-\sin A)[/tex]

substitute the second term in LHS to get

[tex]{ (\cos A - \sin A)^2 \over \cos(2A))}[/tex]

expand the square, [tex] \frac{1-2\sin A \cos A}{\cos 2A}= \sec2A- \tan 2A[/tex]

([tex] \sin (2\theta)=2 \sin\theta\cos\theta[/tex])

Answer:

Both sides can take the same identity

Step-by-step explanation:

Sec 2A - tan 2A

Express with sin and cos

1/cos (2A) - sin (2A)/cos (2A)

= 1 - sin (2A) / cos (2A)

=

1 - sin (x) = 1-sin (x) = 1 - sin^2 ( x ) / 1 + sin (x)  

=

1 - sin ^2 (x) = cos ^2 (x)  

=cos ^2 (2A) / 1+sin (2A)/Cos (2A)

simplified

cos (2A) / 1+sin 2A

identity x used = cos 2x = cos ^2 x - sin ^2 x

= 1 + sin (2x) + (cos (x)) + (sin (x))^2

= cos ^2A - sin ^2A / (cos A - sin A) ^2

= cos A - sin A / cos A +sin A

Both sides are true.

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