Question

trigo........... solution

Answer 1

Answer:

cotA tanB

Step-by-step explanation:

$Given:\frac{cotA+tanB}{cotB+tanA}$

We know that Cot θ = (1/tan θ) (or) tan θ = (1/cotθ)

$=>\frac{cotA+\frac{1}{cotB}}{cotB+\frac{1}{cotA}}$

$=>\frac{\frac{cotBcotA+1}{cotB}}{\frac{cotAcotB+1}{cotA}  }$

$=>\frac{cotA}{cotB}$

$=>\frac{cotA}{\frac{1}{tanB}}$

$=>\boxed{cotAtanB}$

Answer 2

Refer to attachment
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Cot theta=1/tan theta