Question

Prove that :-$(1) \: \frac{tan \theta}{1 - cot \theta} + \frac{cot\theta }{1 - tan \theta} = 1 + tan\theta + cot\theta \: \:$
​

Answer 1

To Prove :-

• $\sf \purple { \: \dfrac{tan\theta}{1 - cot\theta} + \dfrac{cot\theta}{1 - tan\theta} = 1 + tan\theta + cot\theta }\: \:$

Proof :-

L.H.S

$\sf \pink {:\implies \dfrac{tan\theta}{1 - \dfrac{1}{tan\theta} } + \dfrac{ \dfrac{1}{tan\theta} }{1 - tan\theta}} \\ \\ \\ :\implies \sf \dfrac{tan\theta}{ \dfrac{tan\theta - 1}{tan\theta} } + \frac{1}{tan\theta(1 - tan\theta)} \\ \\ \\ :\implies \sf \dfrac{ {tan}^{2} \theta}{tan\theta - 1} - \dfrac{1}{tan\theta(tan\theta - 1)} \\ \\\\ :\implies \sf \dfrac{ {tan}^{3}\theta - 1}{tan\theta(tan\theta - 1)} \\ \\\\ :\implies \sf \dfrac{(tan\theta - 1)( {tan}^{2}\theta + {1}^{2} + tan\theta) }{tan\theta(tan\theta - 1)} \\ \\ \\ :\implies \sf \dfrac{( {tan}^{2}\theta + {1}^{2} + tan\theta) }{tan\theta} \\ \\\\ :\implies \sf \dfrac{ {tan}^{2} \theta}{tan\theta} + \dfrac{1}{tan\theta} + \dfrac{tan\theta}{tan\theta} \\ \\\\ \sf\pink {: \implies tan\theta + cot\theta + 1}$

=R.H.S

• Hence, (Proved..!!)