Question

If cotAcotB = 3 then cos(A+B)/cos(A-B)​

Answer 1

Answer:

1/2

Step-by-step explanation:

[cos(A + B) = cosAcosB - sinAsinB]

[cos(A - B) = cosAcosB + sinAsinB]

Apply the formulas directly in the given equation.

cos(A + B)/cos(A - B)

=> (cosAcosB - sinAsinB)/(cosAcosB + sinAsinB)

Divide the numerator and denominator by sinAsinB

=> (cosAcosB/sinAsinB - sinAsinB/sinAsinB)/(cosAcosB/sinAsinB + sinAsinB/sinAsinB)

=> cotAcotB - 1/cotAcotB + 1

=> 3 - 1/3 + 1

=> 2/4

=> 1/2

Hence, the value of cos(A + B)/cos(A - B) = 1/2.

#Hope my answer help you

Answer 2

$\huge\bf{Answer:-}$

$= cos(A + B) = cosAcosB - sinAsinB \\ </p><p></p><p> = cos(A - B) = cosAcosB + sinAsinB$

Formula

$= cos(A + B)/cos(A - B) \\ </p><p> = \frac{(cosAcosB - sinAsinB)}{(cosAcosB + sinAsinB)}$

$= \frac{ (cosAcosB}{sinAsinB - sinAsinB \frac{(sinAsinB)}{cosAcosB \frac{}sinAsinB + sinAsinB}{sinAsinB)} }$

$= \frac{cotAcotB - 1}{cotAcotB + 1} \\ </p><p> = \frac{3 - 1}{3 + 1} \\ </p><p> = \frac{2}{4} \\ </p><p> = \frac{1}{2} \\ </p><p>$

Therefore, cos(A + B)/cos(A - B) = 1/2