Question

(g) If cos A = sin B = -1/3 . where A lies in 2nd quadrant and B lies in 3rd quadrant, find (tan A - tan B)/(tan A+ tan B).

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Answer 1

Answer:

$1\frac{2}{7}$

Step-by-step explanation:

$cosA=-\frac{1}{3}$

$\frac{x_A}{r_A} =\frac{-1}{3}$ (2nd quadrant → x negatif, y positive)

$r_A^2=x_A^2+y_A^2$

$3^2=(-1)^2+y_A^2$

$y_A=2\sqrt{2}$

$tanA=\frac{y_A}{x_A}$

        $=\frac{2\sqrt{2} }{-1}$

        $=-2\sqrt{2}$

$sinB=-\frac{1}{3}$

$\frac{y_B}{r_B} =\frac{-1}{3}$ (3nd quadrant → x negatif, y negative)

$r_B^2=x_B^2+y_B^2$

$3^2=x_B^2+(-1)^2$

$x_B=-2\sqrt{2}$

$tanB=\frac{y_B}{x_B}$

        $=\frac{-1}{-2\sqrt{2}}$

        $=\frac{1}{4} \sqrt{2}$

$\frac{tanA-tanB}{tanA+tanB}$

$=\frac{-2\sqrt{2}-\frac{1}{4} \sqrt{2} }{-2\sqrt{2}+\frac{1}{4} \sqrt{2} }$

$=\frac{-\frac{9}{4}\sqrt{2} }{-\frac{7}{4} \sqrt{2} }$

$=\frac{9}{7}$

$=1\frac{2}{7}$