Question

pls answer the 3rd question

Answer 1

Continuation...

$= > \frac{2tanA}{1 - tan^2A} +\frac{2tanB}{1 - tan^2B} +\frac{2tanC}{1 - tan^2C} =(\frac{2tanA}{1 - tan^2A})(\frac{2tanB}{1 - tan^2B})(\frac{2tanC}{1 - tan^2C})$

$= >\frac{2x}{1 - x^2} +\frac{2y}{1 - y^2} +\frac{2z}{1 - z^2} =(\frac{2x}{1-x^2})(\frac{2y}{1 - y^2})(\frac{2z}{1 - z^2})$

Hope this helps!