Maths Help Me Please?

Question

Hello,

There are some questions that I am not sure how to do:

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

2) sin2A = (2tanA) / (1 + tan^2A)

3) cos2A = (1 - tan^2A) / (1 + tan^2A)

Thank you in advance.

Answer 1

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

Left hand side identity :(sin2A) / (1 + cos2A)

Double angle formula : sin(2A) = 2sin A cos A,co2A = 2cos²A-1

(sin2A) / (1 + cos2A) = (2sin A cos A) / (1 + 2cos²A-1)

                                           = (2sin A cos A) / 2cos²A

                                = sin A  / cosA

                                = tan A

                                   = Right hand side identity.

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2) sin2A = (2tanA) / (1 + tan^2A)

Left hand side identity :sin2A

Double angle formula : sin(2A) = 2sin A cos A

Sin 2A = 2sinAcosA

Multiply divide by cos A
         = 2 sinA* (cos A)^2/cos A
         = 2 (sinA/cosA)* (cos A)^2
         = 2tanA* 1/(sec x)^2

Pythagorean identity : sec² A - tan² A = 1.
         = 2tan A/(1+ tan^2A)

        = Right hand side identity.

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3) cos2A = (1-tan^2A) / (1 + tan^2A)

Left hand side identity :cos2A

Double angle formula : cos(2A) =cos^2 A-sin^2A

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

Pythagorean identity : sin² A +cos² A = 1.

=(cos^2 A-sin^2A)/(sin² A +cos² A)

Divide numerator and denominator by cos^2 A.

=(cos^2 A-sin^2A)/cos^2 A/(sin² A +cos² A)/cos^2 A

=(1-tan^2A)/(1+tan^2A)

= Right hand side identity.