Find the exact value under the given conditions.
-$\cos \alpha = - \frac { 24 } { 25 } , \frac { \pi } { 2 }

Question

Find the exact value under the given conditions.
-$\cos \alpha = - \frac { 24 } { 25 } , \frac { \pi } { 2 } < \alpha < \pi ; \quad \sin \beta = - \frac { \sqrt { 21 } } { 5 } , \pi < \beta < \frac { 3 \pi } { 2 } \quad$ Find $\tan ( \alpha + \beta )$
A) $\frac { - 14 + 24 \sqrt { 21 } } { 48 + 7 \sqrt { 21 } }$
B) $\frac { 48 + 7 \sqrt { 21 } } { - 14 + 24 \sqrt { 21 } }$
C) $\frac { 14 - 24 \sqrt { 21 } } { 48 + 7 \sqrt { 21 } }$
D) $\frac { - 14 + 24 \sqrt { 21 } } { 125 }$

Quizplus · Accepted Answer

Given $\cos \alpha = - \frac { 24 } { 25 }$ and $\sin \beta = - \frac { \sqrt { 21 } } { 5 }$, we find $\sin \alpha$ and $\cos \beta$ using Pythagoras' theorem. For $\alpha$, $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left(-\frac{24}{25}\right)^2} = \frac{7}{25}$ (positive because $\frac{\pi}{2} < \alpha < \pi$). For $\beta$, $\cos \beta = -\sqrt{1 - \sin^2 \beta} = -\sqrt{1 - \left(-\frac{\sqrt{21}}{5}\right)^2} = -\frac{2}{5}$ (negative because $\pi < \beta < \frac{3\pi}{2}$).Using the formula for $\tan(\alpha + \beta) = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta}$, we substitute the values:$$\tan(\alpha + \beta) = \frac{\frac{7}{25} \cdot \left(-\frac{2}{5}\right) + \left(-\frac{24}{25}\right) \cdot \left(-\frac{\sqrt{21}}{5}\right)}{\left(-\frac{24}{25}\right) \cdot \left(-\frac{2}{5}\right) - \frac{7}{25} \cdot \left(-\frac{\sqrt{21}}{5}\right)} = \frac{-\frac{14}{125} + \frac{24\sqrt{21}}{125}}{\frac{48}{125} + \frac{7\sqrt{21}}{125}} = \frac{-14 + 24\sqrt{21}}{48 + 7\sqrt{21}}$$Thus, the correct answer is A) $\frac{-14 + 24\sqrt{21}}{48 + 7\sqrt{21}}$.

Find the Exact Value Under the Given Conditions Find tan⁡(α+β)\tan ( \alpha + \beta )tan(α+β)

Find the Exact Value Under the Given Conditions Find $\tan ( \alpha + \beta )$