Question

If cos A = cos B = $-\frac{1}{2}$ and A does not lie in the second quadrant and B does not lie in the third quadrant, then find the value of $\frac{4 sin B - 3 tan A}{tan B + sin A}$

Answer 1

we know, cosx be negative if x lies in either 2nd quadrant or 3rd quadrant.

A/C to question, If cos A = cos B = $-\frac{1}{2}$ and A does not lie in the second quadrant and B does not lie in the third quadrant.
it means, A lies in 3rd quadrant and B lies in 2nd quadrant.

now, cosA = -1/2
sinA = ±√3/2 , but A lies in 3rd quadrant
so, sinA = -√3/2 .
similarly, tanA = √3 [ tan will be positive in 3rd quadrant ]

now, cosB = -1/2
sinB = ±√3/2 , but B lies in 2nd quadrant
so, sinB ≠ -√3/2 hence, sinB = √3/2
and tanB = -√3 [ tan is negative in 2nd quadrant]

now, (4sinB - 3tanA)/(tanB + sinA)

= (4 × √3/2 - 3 × √3)/(-√3 -√3/2)

= (2√3 - 3√3)/{(-3√3)/2}

= (-√3)/(-3√3/2)

= 2/3

Answer 2

Answer:2/3

Step-by-step explanation:

it is given that

cos A =cos B=-1/2 --(1)

A does not lie on Q2 and B does not lie in O3

So,A lies in Q3 and B lies in Q2

From Eq 1

Sin B =$root$ 3/2

Tan A =root 3

Tan B is negitive because b lies in Q2

Tan B =-Root 3

Sin A=$root$ 3

So,after substituting the values answer is 2/3