Question

$(\frac{tan \: a}{1 - cot \: a} ) + ( \frac{cot \: a}{1 - tan \: a}) = (1 + tan \: a + cot \: a)$
please answer my question​

Answer 1

$\bf LHS = \left( \dfrac{tan \: a}{1 - cot \: a} \right ) + \left (\dfrac{cot \: a}{1 - tan \: a} \right) \\ \\ \bf\left( \dfrac{tan \: a}{1 - \frac{1}{tan \: a}} \right ) + \left (\dfrac{ \frac{1}{tan \: a} }{1 - tan \: a} \right) \\ \\ \bf\left( \dfrac{tan \: a}{ \dfrac{tan \: a- 1}{tan \: a}} \right ) + \left (\dfrac{ \frac{1}{tan \: a} }{1 - tan \: a } \right) \\ \\ \bf\left( \dfrac{tan^{2} \: a}{tan \: a- 1} \right ) + \left (\dfrac{ 1 }{tan \: a(1 - tan \: a)} \right) \\ \\ \bf\left( \dfrac{tan^{3} \: a - 1}{tan \: a(tan \: a- 1)} \right ) \\ \\ { \boxed{ \pmb{ \sf{ \red{ {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )}}}}} \\ \\ \bf\left( \dfrac{(tan \: a - 1)(tan^{2}a + tan \: a + 1 }{tan \: a( tan \: a - 1)} \right) \\ \\ \bf\left( \dfrac{tan^{2}a + tan \: a + 1 }{tan \: a} \right) \\ \\ \bf \dfrac{tan^{2}a}{tan \: a} + \dfrac{tan \: a}{tan \: a} + \dfrac {1 }{tan \: a} \\ \\ \bf tana+ 1+ cot \: a = RHS\: verified$

Quick Review :

• Convert cot a into tan a
• cot a = 1/tan a