Question

if A is not in 2nd quadrant,B is not in 3rd quadrant and cos A=-1/2,cos B=-1/2,then find the value of 4 sinB-3 tanA/tanB+sinA

Answer 1

Answer:

$\frac{2}{3}$

Step-by-step explanation:

We have,

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

where A is not in 2nd quadrant,B is not in 3rd quadrant.

$cos(A)=\frac{-1}{2}\\ \\ cos(A)=cos(\pi+\pi/3)\\ \\A=4\pi/3$

Hence, this value of A lies in 3rd Quadrant.

Also,

$cos(B)=\frac{-1}{2}\\ \\ cos(B)=cos(\pi-\pi/3)\\ \\B=2\pi/3$

Ths value of B lies in 2nd Quadrant.

Now,

$\frac{4 sinB-3 tanA}{tanB+sinA}=\frac{4 \frac{\sqrt3}{2}-3 *\sqrt3}{-\sqrt{3}-\frac{\sqrt{3}}{2}}\\ \\ =\frac{2\sqrt{3}-3\sqrt{3}}{\frac{-3\sqrt{3}}{2}}\\ \\=\frac{2}{3}$

Answer 2

Answer:

2/3

Step-by-step explanation:

Solution

Consider the given data.

cos

A

=

cos

B

=

−

1

2

We know that cosine function is negative in second quadrant and third quadrant, so if A does not lie in 2nd quadrant, it must i.e. in 3rd quadrant $$ if B does not lie 3rd quadrant, it must lie in 2nd quadrant now, we have

cos

A

=

−

1

2

cos

A

=

−

cos

π

3

⇒

cos

(

π

+

π

3

)

cos

A

=

cos

(

4

π

3

)

A

=

4

π

3

Similarly,

cos

B

=

−

1

2

cos

B

=

−

cos

π

3

⇒

cos

(

π

−

π

3

)

cos

B

=

cos

(

2

π

3

)

B

=

2

π

3

Therefore,

sin

A

=

sin

4

π

3

=

−

√

3

2

tan

A

=

tan

4

π

3

=

√

3

sinB

=

sin

2

π

3

=

√

3

2

tanB

=

tan

2

π

3

=

−

√

3

Since,

4

sinB

−

3

tan

A

tan

B

+

sin

A

=

4

×

√

3

2

−

3

×

√

3

−

√

3

−

√

3

2

=

2

√

3

−

3

√

3

−

3

√

3

2

=

−

√

3

−

3

√

3

2

=

2

3

Hence, the value is

2

3

.