Question

Please solve this. Maths lover I need your help!

Answer 1

$\sf{Given :\;\dfrac{Tan\theta}{1 - Cot\theta} + \dfrac{Cot\theta}{1 - Tan\theta}}\\\\\\\boxed{\sf{We\;know\;that : Cot\theta = \dfrac{1}{Tan\theta}}}\\\\\\\implies \dfrac{Tan\theta}{1 - \dfrac{1}{Tan\theta}} + \dfrac{\dfrac{1}{Tan\theta}}{1 - Tan\theta}\\\\\\\implies \dfrac{Tan\theta}{\dfrac{Tan\theta - 1}{Tan\theta}} + \dfrac{1}{Tan\theta(1 - Tan\theta)}\\\\\\\implies \dfrac{Tan^2\theta}{(Tan\theta - 1)} - \dfrac{1}{Tan\theta(Tan\theta - 1)}$

$\sf{\implies \dfrac{1}{(Tan\theta - 1)}\times (Tan^2\theta - \dfrac{1}{Tan\theta}})}\\\\\\\sf{\implies \dfrac{1}{(Tan\theta - 1)}\times (\dfrac{Tan^3\theta - 1}{Tan\theta})$

$\boxed{\sf{We\;know\;that : a^3 - b^3 = (a - b)(a^2 + ab + b^2)}}\\\\\\\sf{\implies \dfrac{1}{(Tan\theta - 1)}\times [\dfrac{(Tan\theta - 1)(Tan^2\theta + Tan\theta + 1)}{Tan\theta}]$

$\sf{\implies \dfrac{Tan^2\theta + Tan\theta + 1}{Tan\theta}$

$\sf{\implies 1 + Tan\theta + \dfrac{1}{Tan\theta}}\\\\\\\sf{\implies 1 + \dfrac{Tan^2\theta + 1}{Tan\theta}$

$\boxed{\sf{We\;know\;that : 1 + Tan^2\theta = Sec^2\theta}}$

$\sf{\implies 1 + \dfrac{Sec^2\theta}{Tan\theta}}\\\\\\\sf{\implies 1 + \dfrac{Sec^2\theta.Cos\theta}{Sin\theta}$

$\boxed{\sf{We\;know\;that : Sec\theta.Cos\theta = 1}}$

$\sf{\implies 1 + \dfrac{Sec\theta}{Sin\theta}$

$\boxed{\sf{We\;know\;that : \dfrac{1}{Sin\theta} = Cosec\theta}}}$

$\sf{\implies 1 + Sec\theta.Cosec\theta}$