Question

Prove the following

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

Answer 1

Tan divided by cot in terms of perpendicular and base is: (p/b) / (b/p) = p*p / b*b . Tan divided by cot in terms of sin and cos is : sin*sin / cos*cos

Answer 2

Step-by-step explanation:

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