Question

plz answer with full explanation....​

Answer 1

$\bold{ \large{ \underline{Formula }}}$

$\large \to \bold{\tan(a + b) = \frac{ \tan(a) + \tan(b) }{1 - \tan(a) \: \times \: \tan(b) } } \\ \\ \large \to \bold{ \tan(a - b) = \frac{ \tan(a) - \tan(b) }{1 + \tan(a) \: \times \: \tan(b) } }$

$\bold{ \large{ \underline{RHS \: } }}$

$\large \to \bold{ { ( \: \frac{1 + \tan(x) }{1 - \tan(x) } \: )}^{2} } \:$

$\bold{ \large{ \underline{LHS } }}</p><p> \:$

$\large \to\bold{ \frac{\tan( \frac{\pi}{4} + x ) }{ \tan( \frac{\pi}{4} - x) } }\\ \\ \large \to \bold{\frac{ \frac{ \tan( \frac{\pi}{4} ) + \tan(x) }{1 - \tan( \frac{\pi}{4}) \tan(x) } } { \frac{ \tan( \frac{\pi}{4} )- \tan(x) }{ 1 + \tan( \frac{\pi}{4} ) \tan(x) } } } \\ \\ \large \to \bold{ \frac{ \frac{1 + \tan(x) }{1 - \tan(x) } }{ \frac{1 - \tan(x) }{1 + \tan(x) } } } \\ \\ \large \to \bold{\frac{(1 + \tan(x))(1 + \tan(x) )}{(1 - \tan(x))(1 - \tan(x) )}}\\ \\ \large \to \bold{ { ( \: \frac{1 + \tan(x) }{1 - \tan(x) } \: )}^{2} }$

$\bold{ \large{ \underline{ \underline{ \: \: RHS \: = \: LHS \: , \: Hence \: Proved \: \: \: }}}}$

Answer 2

rhs = lhs

hence proved

thank you