Sin-Tanx Error!!!

$sin2\theta =\cfrac { 2tan\theta }{ 1+{ tan }^{ 2 }\theta } =\cfrac { 2*2 }{ 1+4 } =4/5=\boxed { 0.8 }$

tan@ = 2 Can be true for a right angled triangle with

P = 2, B= 1 and H = sqrt(5)

Hence.

             sin@ = 2/sqrt(5)

               cos@ = 1 /sqrt(5)

and

              sin2@ = 2sin@cos@

                              = 2(2/sqrt(5))(1/sqrt(5))

                               =4/5 =0.8

Note: symbol @ is used for'theta'.

$\tan{\theta} = 2 \\ \tan^2{\theta} = 4 \\ \sec^2{\theta} = 1 + tan^2{\theta} = 1 + 4 = 5 \\ \cos^2{\theta} = \frac{1}{5} \\ \sin^2{\theta} = 1 - cos^2{\theta} = \frac{4}{5} \\ \sin{2\theta} = 2\cos{\theta}\sin{\theta} \\ sin^2{2\theta} = 4\cos^2{\theta}\sin^2{\theta} \\ \sin^2{2\theta} = 4 *\frac{4}{5}*\frac{1}{5} = \frac{16}{25} \\ \boxed{\sin{2\theta} = \frac{4}{5}}$