Question

It’s PreCal I just need a method to solve it

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using the identities

tanx = $\frac{sinx}{cosx}$ , cotx = $\frac{cosx}{sinx}$

Consider the left side

$\frac{tanx}{tanx+cotx}$

= $\frac{\frac{sinx}{cosx} }{\frac{sinx}{cosx}+\frac{cosx}{sinx} }$

= $\frac{\frac{sinx}{cosx} }{\frac{sin^2x+cos^2x}{cosxsinx} }$

= $\frac{\frac{sinx}{cosx} }{\frac{1}{cosxsinx} }$

= $\frac{sinx}{cosx}$ × $\frac{cosxsinx}{1}$ ← cancel cosx on numerator/ denominator

= sin²x

= right side