Question

Tan(45+alpha)+tan(45-alpha)/cot(45+alpha)+cot(45-alpha)

Answer 1

hope this will help you enjoy

Answer 2

➡HERE IS YOUR ANSWER⬇

First Process :

$\frac{tan(45 + \alpha ) + tan(45 - \alpha )}{cot(45 + \alpha ) + cot(45 - \alpha )} \\ \\ = \frac{ \frac{tan45 + tan \alpha }{1 - tan45 \: tan \alpha } + \frac{tan45 - tan \alpha }{1 + tan45 \: tan \alpha } }{ \frac{cot45 \: cot \alpha - 1}{ cot \alpha + cot45 } + \frac{cot45 \: cot \alpha + 1}{cot \alpha - cot45} } \\ \\ = \frac{ \frac{1 + tan \alpha }{1 - tan \alpha } + \frac{1 - tan \alpha }{1 + tan \alpha } }{ \frac{cot \alpha - 1}{cot \alpha + 1} + \frac{cot \alpha + 1}{cot \alpha - 1} } \\ \\ = \frac{ \frac{(1 + tan \alpha )(1 + tan \alpha ) + (1 - tan \alpha )(1 - tan \alpha )}{(1 - tan \alpha )(1 + tan \alpha )} }{ \frac{(cot \alpha - 1 )(cot \alpha - 1) + (cot \alpha + 1)(cot \alpha + 1)}{(cot \alpha + 1)(cot \alpha - 1)} } \\ \\ = \frac{ \frac{1 + 2tan \alpha + {tan}^{2} \alpha + 1 - 2tan \alpha + {tan}^{2} \alpha }{1 - {tan}^{2} \alpha } }{ \frac{ {cot}^{2} \alpha - 2cot \alpha + 1 + {cot}^{2} \alpha + 2cot \alpha + 1 }{ {cot}^{2} \alpha - 1 } } \\ \\ = \frac{ \frac{2(1 + {tan}^{2} \alpha )}{1 - {tan}^{2} \alpha } }{ \frac{2( {cot}^{2} \alpha + 1)}{ {cot}^{2} \alpha - 1 } } \\ \\ = \frac{1 + {tan}^{2} \alpha }{1 - {tan}^{2} \alpha } \times \frac{ {cot}^{2} \alpha - 1 }{ {cot}^{2} \alpha + 1} \\ \\ = \frac{1 + {tan}^{2} \alpha }{1 - {tan}^{2} \alpha } \times \frac{ \frac{1}{ {tan}^{2} \alpha } - 1}{ \frac{1}{ {tan}^{2} \alpha } + 1} \\ \\ = \frac{1 + {tan}^{2} \alpha }{1 - {tan}^{2} \alpha } \times \frac{1 - {tan}^{2} \alpha }{1 + {tan}^{2} \alpha } \\ \\ = 1$

Second Process :

$\frac{tan(45 + \alpha ) + tan(45 - \alpha )}{cot(45 + \alpha ) + cot(45 - \alpha )}\\ \\=\frac{tan(45 + \alpha ) + tan(45 - \alpha )}{cot(90-(45 - \alpha )) + cot(90-(45 + \alpha) )}\\ \\= \frac{tan(45 + \alpha ) + tan(45 - \alpha )}{tan(45 - \alpha ) + tan(45 + \alpha)}\\ \\= 1$

⬆HOPE THIS HELPS YOU⬅