Question

value of 1-tan^2 50°/1+tan^2 50°​

Answer 1

SOLUTION.

$\frac{1 - {tan}^{2} 50}{1 + {tan}^{2}50 } = y \\ \\ \frac{1 - \frac{ {sin}^{2}50 }{ {cos}^{2}50 } }{1 + \frac{ {sin}^{2}50 }{ {cos}^{2}50 } } = y \\ \\ \frac{ \frac{ {cos}^{2}50 - {sin}^{2}50 }{ {cos}^{2}50 } }{ \frac{ {sin}^{2}50 + {cos}^{2}50 }{ {cos}^{2}50 } } = y \\ \\ y = {cos}^{2} 50 - {sin}^{2} 50 \\ \\ y = \cos2(50) \\ \\ y = \cos100 \\ \\ y = - 0.17 \\ \\ which \: is \: not \: exact \: but \: the \: actual \: value \: is \: close \: to \: it$

CONCEPT USED.

$\cos2( \alpha ) = {cos}^{2} \alpha - {sin}^{2} \alpha \\ \\ \tan( \alpha ) = \frac{sin \alpha }{cos \alpha }$