Question

tan theta /1 - cot theta +cot theta /1 - tan theta =1 +tan theta +cot theta :prove it

Answer 1

Step-by-step explanation:

Please check the question  .

The possible answer is as follows

tanθ/1- cotθ  + cotθ/1-tanθ = 1+tanθ+cotθ

Let us take L.H.s

tanθ/1- cotθ  + cotθ/1-tanθ

=[sinθ/cosθ/ 1- (cosθ/sinθ)] + [cosθ/1- (sinθ/cosθ)]

=sin²θ/cosθ( sinθ-cosθ) + cos²θ/ sinθ(sinθ-cosθ)

=sin³θ -cos³θ/cosθsinθ(sinθ-cosθ)

∴a³-b³= (a-b)[a²+b²+ab]

=(sinθ-cosθ)[sin²θ+cos²θ+sinθcosθ]/cosθsinθ(sinθ-cosθ)

=1+ sinθcosθ/cosθsinθ

= 1+secθcosecθ≠L.H.S

hence question should be tanθ/1- cotθ  + cotθ/1-tanθ =1+secθcosecθ

Answer 2

$LHS=\frac{tan\theta}{1-cot\theta}+\frac{cot\theta}{1-tan\theta}\\\\=\frac{tan\theta}{1-\frac{1}{tan\theta}}+\frac{cot\theta}{1-tan\theta}\\\\=\frac{tan^2\theta}{tan\theta-1}-\frac{cot\theta}{tan\theta-1}\\\\=\frac{tan^2\theta-cot\theta}{tan\theta-1}\\\\=\frac{tan^3\theta-1}{tan\theta(tan\theta-1)}\\\\=\frac{(tan\theta-1)(tan^2\theta+tan\theta+1)}{tan\theta(tan\theta-1)}\\\\=\frac{tan^2\theta+tan\theta+1}{tan\theta}\\\\=\frac{tan^2\theta}{tan\theta}+\frac{tan\theta}{tan\theta}+\frac{1}{tan\theta}\\\\=tan\theta+1+cot\theta\\\\=1+tan\theta+cot\theta=RHS$