Question

if cos A= 0.5 and cos B=1/under root 2 :find the value of tanA -tan B/1+tan A tan B​

Answer 1

Answer:

Step-by-step explanation:

in right triangle ABC, let A & B be the non right angles

let cosA = $\frac{5}{10}$

therefore hypo = 10

AC = 5

since CosB = $\frac{1}{\sqrt{2} }$

therefore BC = $\sqrt{75}$ (from pythagoras theorem)

from these sides,

TanA=$\frac{\sqrt{75}}{5}$ = $\sqrt{3}$

TanB=$\frac{5}{\sqrt{75}}$ = $\frac{1}{\sqrt{3} }$

substituting in question, we get:

$\frac{(\sqrt{3}-\frac{1}{\sqrt{3} }) }{1+(\sqrt{3}*\frac{1}{\sqrt{3} }) }$

Therefore the answer is $(\frac{1}{\sqrt{3} }) }$