Question

Prove that tano/1-coto+coto/1-tano =1+tano+coto​

Answer 1

Given to proof:-

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

Proof:-

Considering $\underline{\boxed{\sf{\pink{L.H.S.}}}}$,

$\sf \dfrac{tan\theta}{1 - cot\theta} + \dfrac{cot\theta}{1 - tan\theta}$

• We know that, cotθ = 1/tanθ

$\sf \longrightarrow \Bigg \{\dfrac{tan\theta}{ \bigg(1 - \dfrac{1}{tan\theta} \bigg) } \Bigg \} + \Bigg \{\dfrac{ \Big(\dfrac{1}{tan\theta} \Big) }{1 - tan\theta} \Bigg \}$

$\sf \longrightarrow \Bigg \{\dfrac{tan\theta}{ \bigg( \dfrac{tan\theta - 1}{tan\theta} \bigg) } \Bigg \} + \dfrac{1}{tan\theta(1 - tan\theta)}$

$\sf \longrightarrow \dfrac{tan { }^{2}\theta }{tan\theta - 1} + \dfrac{1}{tan\theta(1 - tan\theta )}$

• Multiplying numerator and denominator by (-1) in first term and then rearranging,

$\sf \longrightarrow \dfrac{1}{tan\theta(1 - tan\theta )} - \dfrac{tan { }^{2} \theta}{1 - tan\theta}$

$\sf \longrightarrow \dfrac{1 - tan { }^{3}\theta }{tan\theta(1 - tan\theta)}$

• Expanding numerator using a³ - b³ = (a - b)(a² + b² + ab),

$\sf \longrightarrow \dfrac{(1 - tan\theta)(1 + tan { }^{2} \theta + tan\theta)}{tan\theta(1 - tan\theta)}$

$\sf \longrightarrow \dfrac{ \cancel{(1 - tan\theta)}(1 + tan { }^{2} \theta + tan\theta)}{tan\theta \cancel{(1 - tan\theta)}}$

$\sf \longrightarrow \dfrac{1 + {tan}^{2}\theta + tan\theta }{tan\theta}$

$\sf \longrightarrow \dfrac{1 }{tan\theta} + \dfrac{ {tan}^{2}\theta }{ tan\theta} + \dfrac{tan\theta}{tan\theta}$

$\sf \longrightarrow cot\theta + tan\theta + 1$

$\longrightarrow \underline{\underline{\sf{\green{1 + tan\theta + cot\theta}}}}$

$\leadsto \underline{\boxed{\sf{\pink{ R.H.S. }}}}$

So, L.H.S. = R.H.S.

Hence verified.