Question

find the value of tan(alpha+beta), given that cot alpha=1/2, alpha€(pi,3pie/2), sec beta=-5/3, beta€(pie/2,pie). Also state the quadrant in which alpha+beta terminates

Answer 1

Answer:

$tan(\alpha+\beta)=\frac{2}{11}$

Step-by-step explanation:

Given:

$cot\alpha=\frac{1}{2}$ and $sec\beta=\frac{-5}{3}$

$cot\alpha=\frac{1}{2}$

Taking reciprocals

$tan\alpha=2$ and

$sec\beta=\frac{-5}{3}$

$tan^2\beta=sec^2\beta-1$

$tan^2\beta=(\frac{-5}{3})^2-1$

$tan^2\beta=\frac{25}{9}-1$

$tan^2\beta=\frac{25-9}{9}$

$tan^2\beta=\frac{16}{9}$

$tan\beta=\sqrt{\frac{16}{9}}$

$tan\beta=$±$\frac{4}{3}$

since $\beta$ lies in II quadrant, $tan\beta$ is negative

$tan\beta=\frac{-4}{3}$

Now,

$tan(\alpha+\beta)$

$=\frac{tan\alpha+tan\beta}{1-tan\alpha.tan\beta}$

$=\frac{2+(\frac{-4}{3})}{1-2(\frac{-4}{3})}$

$=\frac{\frac{6-4}{3}}{1+\frac{8}{3}}$

$=\frac{\frac{6-4}{3}}{1+\frac{8}{3}}$

$=\frac{\frac{2}{3}}{\frac{3+8}{3}}$

$=\frac{\frac{2}{3}}{\frac{11}{3}}$

$=\frac{2}{11}$

This implies,

$tan(\alpha+\beta)=\frac{2}{11}$

since $tan(\alpha+\beta)$ is positive, $\alpha+\beta$ lies in either I quadrant or III quadrant.

As per data,

$\alpha+\beta$ lies in I quadrant

Answer 2

Answer:

your answer..............

Step-by-step explanation:

tan(α+β)=

11

2

Step-by-step explanation:

Given:

cot\alpha=\frac{1}{2}cotα=

2

1

and sec\beta=\frac{-5}{3}secβ=

3

−5

cot\alpha=\frac{1}{2}cotα=

2

1

Taking reciprocals

tan\alpha=2tanα=2 and

sec\beta=\frac{-5}{3}secβ=

3

−5

tan^2\beta=sec^2\beta-1tan

2

β=sec

2

β−1

tan^2\beta=(\frac{-5}{3})^2-1tan

2

β=(

3

−5

)

2

−1

tan^2\beta=\frac{25}{9}-1tan

2

β=

9

25

−1

tan^2\beta=\frac{25-9}{9}tan

2

β=

9

25−9

tan^2\beta=\frac{16}{9}tan

2

β=

9

16

tan\beta=\sqrt{\frac{16}{9}}tanβ=

9

16

tan\beta=tanβ= ±\frac{4}{3}

3

4

since \betaβ lies in II quadrant, tan\betatanβ is negative

tan\beta=\frac{-4}{3}tanβ=

3

−4

Now,

tan(\alpha+\beta)tan(α+β)

=\frac{tan\alpha+tan\beta}{1-tan\alpha.tan\beta}=

1−tanα.tanβ

tanα+tanβ

=\frac{2+(\frac{-4}{3})}{1-2(\frac{-4}{3})}=

1−2(

3

−4

)

2+(

3

−4

)

=\frac{\frac{6-4}{3}}{1+\frac{8}{3}}=

1+

3

8

3

6−4

=\frac{\frac{6-4}{3}}{1+\frac{8}{3}}=

1+

3

8

3

6−4

=\frac{\frac{2}{3}}{\frac{3+8}{3}}=

3

3+8

3

2

=\frac{\frac{2}{3}}{\frac{11}{3}}=

3

11

3

2

=\frac{2}{11}=

11

2

This implies,

tan(\alpha+\beta)=\frac{2}{11}tan(α+β)=

11

2

since tan(\alpha+\beta)tan(α+β) is positive, \alpha+\betaα+β lies in either I quadrant or III quadrant.

As per data,

\alpha+\betaα+β lies in I quadrant