If A = 300 , verify that: (i) sin 2A = 2 tanA/1+ tan^2A

Question

If A = 30°, verify that:

(i) sin 2A = \(\frac{2tanA}{1+tan^2A}\)

(ii) cos 2A = \(\frac{1-tan^2A}{1+tan^2A}\)

(iii) tan 2A = \(\frac{2tanA}{1-tan^2A}\)

Answer 1

A = 30°

⇒ 2A = 2 × 30° = 60°

(i) sin 2A = sin 60° = √3/2

\(\frac{2tanA}{1+tan^2A}\) = \(\frac{2tan30°}{1+tan^230°}\)

∴ sin 2A = \(\frac{2tanA}{1+tan^2A}\)

(ii) cos 2A = cos 60° = 1/2

\(\frac{1-tan^2A}{1+tan^2A}\) = \(\frac{1-tan^230°}{1+tan^230°}\)

∴ cos 2A = \(\frac{1-tan^2A}{1+tan^2A}\)

(iii) tan 2A = tan 60° = √3

\(\frac{2tanA}{1-tan^2A}\) = \(\frac{2tan30°}{1-tan^230°}\)

∴ tan 2A = \(\frac{2tanA}{1-tan^2A}\)