Question

Given cos(α) = -3/14, with α in quadrant III, and sin(β) = -12/23, with β in quadrant IV. Find tan(α + β)

Answer 1

(√71295 - 36) / (3 + 12√187)

Answer 2

is it correct..?

Answer 3

This is a long answer explaining everything that you may need, your desired/final answer is at the bottom.

We can begin by finding the value of sin(α) using the Pythagorean identity sin^2(α) + cos^2(α) = 1.
Since cos(α) = -3/14, we have (-3/14)^2 + sin^2(α) = 1.
Simplifying, we get 9/196 + sin^2(α) = 1.
Subtracting 9/196 from both sides we find sin^2(α) = 187/196.
Taking the square root of both sides and remembering that α is in quadrant III (where sine is negative), we find sin(α) = -√(187/196) = -√187/14.

Next, we need to find the value of cos(β). Since sin(β) = -12/23 and β is in quadrant IV, we can use the Pythagorean identity sin^2(β) + cos^2(β) = 1.
Plugging in sin(β) = -12/23, we have (-12/23)^2 + cos^2(β) = 1.
Simplifying, we get 144/529 + cos^2(β) = 1.
Subtracting 144/529 from both sides, we find cos^2(β) = 385/529.
Taking the square root of both sides and remembering that β is in quadrant IV (where cosine is positive), we find cos(β) = √(385/529) = √385/23.

Finally, we can use the formula for tan(α + β) = (tan(α) + tan(β))/(1 - tan(α)tan(β)).
Since tan(α) = sin(α)/cos(α) and tan(β) = sin(β)/cos(β), we can substitute the known values.
We find tan(α + β) = (sin(α)/cos(α) + sin(β)/cos(β))/(1 - sin(α)/cos(α) * sin(β)/cos(β)).
Substituting the values we found earlier, we have tan(α + β) = (-√187/14)/(√385/23)/(1 - (-√187/14)(-√385/23)).
Multiplying the numerator and the denominator by the reciprocal of the denominator, we have tan(α + β) = (-√187/14) * (23/√385) / (1 - (187/14)*(385/23)).
Simplifying the fraction, we have tan(α + β) = (-23√187/(14√385)) / (1 - 385/14) = (-23√187/(14√385)) / (14/14 - 385/14).
Simplifying the denominator, we have tan(α + β) = (-23√187/(14√385)) / (-371/14).
Dividing by a fraction is equivalent to multiplying by the reciprocal, so we get tan(α + β) = (-23√187/(14√385)) * (-14/371).
Simplifying further, we have tan(α + β) = 23√187/(371√385).

Therefore, tan(α + β) = 23√187/(371√385).

Answer 4

Don't know fr

Answer 5

\[\alpha ~\in~quadrant ~|||\]
\[Hence~\cos \alpha~and~\sin \alpha~are~negative.\]
\[\sin \alpha=-\sqrt{1-\cos ^2\alpha}=-\sqrt{1-(\frac{ -3 }{ 14 })^2}\]
\[=-\sqrt{1-\frac{ 9 }{ 196}}=-\sqrt{\frac{ 187 }{ 196 }}\]

\[\beta ~in~quadrant ~iv \]
Therefore cos beta is positive.
\[\cos \beta=-\sqrt{1-\sin ^2\beta}=\sqrt{1-(\frac{ -12 }{ 23 })^2}\]
\[=\sqrt{1-\frac{ 144 }{529 }}=\sqrt{\frac{ 529-144 }{ 529 }}=\sqrt{\frac{ 385 }{ 529 }}\]
\[\tan \alpha=\frac{ \sin \alpha }{ \cos \alpha }=\frac{ \frac{ -\sqrt{187}}{ 14} }{ \frac{ -3 }{ 14 } }=\frac{ \sqrt{187} }{ 3 }\]
\[\tan \beta=\frac{ \sin \beta }{ \cos \beta } =\frac{ \frac{ -12}{ 23 } }{ \frac{ \sqrt{385} }{ 23 } }=-\frac{ 12 }{\sqrt{385}}\]
\[\tan (\alpha+\beta)=\frac{ \tan \alpha+\tan \beta }{ 1-\tan \alpha \tan \beta }\]
\[=\frac{ \frac{ \sqrt{187} }{ 3 }+\frac{ -12 }{ \sqrt{385}} }{ 1-\frac{ \sqrt{187} }{ 3 }\times \frac{ -12 }{ \sqrt{385}} }\]
\[=\frac{ \frac{ \sqrt{187} \times \sqrt{385}-36}{ 3\sqrt{385} } }{ 1+\frac{ 12\sqrt{187} }{ 3\sqrt{385} } }\]
\[=\frac{ \sqrt{71995}-36 }{ 3\sqrt{385}+12\sqrt{187} }\]