Question

1-tanA/1+tanA = 1-sin2A/cos2A​

Answer 1

$\huge\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Answer}}}}}}}} \\ \large\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Your~answer↓}}}}}}}}$

To prove :-

$\boxed{ \large{ { \frac{1 - tanA}{1 + tanA} = \frac{1 - sin2A}{cos2A} }}}$

Formula's used

$\boxed{ \large{ \mathfrak{sin2z = 2sinzcosz}}}$

$\boxed{ \large{ \mathfrak{1 - cos2z = 2 {sin}^{2} z}}}$

Solution :-

Consider LHS :-

$\frac{1 - sin2A}{cos2A} \\ = \frac{1 - cos( \frac{\pi}{2} - 2A)}{sin( \frac{\pi}{2} - 2A)} \\ \small\bold\red{(let \: \frac{\pi}{2} - 2A = y)} \\ so \: above \: can \: be \: rewritten \: as \\ = \frac{1 - cosy}{siny} \\ = \frac{ {2sin}^{2} \frac{y}{2} }{2sin \frac{y}{2}cos \frac{y}{2} } \\ = \frac{sin \frac{y}{2} }{cos \frac{y}{2} } \\ = tan \frac{y}{2} \\ = tan( \frac{ \frac{\pi}{2} - 2A}{2} ) \\ = tan( \frac{\pi}{4} - A) \\ = \frac{1 - tanA}{1 + tanA}$

$\boxed{ \large{ \mathfrak{hence \: proved}}}$

$\boxed{ \large{ \mathfrak{Mark \: me \: as \: Brainliest}}}$