Question

Pleasensolve this proving question​

Answer 1

To prove :

$\star \: \frac{ \tan( \frac{\pi}{4} + x)}{ \tan( \frac{\pi}{4} - x) } = {( \frac{1 + \tan \: x}{ 1 - \tan \: x}) }^{2}$

Solution :

LHS :

$\implies \: \frac{ \tan \: ( \frac{\pi}{4} + x) }{ \tan \: ( \frac{\pi}{4} - x) } \\ \\ \implies \: \frac{ \frac{ \tan \: \frac{\pi}{4} + \tan \: x }{1 - \tan \: \frac{\pi}{4} \tan \: x } }{ \frac{ \tan \: \frac{\pi}{4} - \tan \: x }{1 + \tan \: \frac{\pi}{4} \tan \: x \: } } \\ \\ \implies \: \frac{ \frac{1 + \tan \: x}{1 - \tan \: x} }{ \frac{1 - \tan \: x}{1 + \tan \: x} } \\ \\ \implies \: \frac{1 + \tan \: x}{1 - \tan \: x} \times \frac{1 + \tan \: x}{ 1 - \tan \: x \: } \\ \\ \implies \: \frac{ {(1 + \tan \: x)}^{2} }{ {(1 - \tan \: x)}^{2} } \\ \\ \implies \: {( \frac{1 + \tan \: x}{1 - \tan \: x} )}^{2}$

LHS = RHS

Proved :

Formula :

$\star \bold{\tan \: (x + y) = \frac{ \tan \: x + \tan \: y}{1 - \tan \: x \tan \: y} }\\ \\ \star \: \bold{ \tan(x - y) = \frac{ \tan \: x - \tan \: y}{1 + \tan \: x \tan \: y \: } }$