Answer:
tan(a + B) = [tex]\frac{4}{3}[/tex]
Step-by-step explanation:
cot(a) = [tex]-\frac{3}{4}[/tex]
Therefore, tan(a) = [tex]-\frac{4}{3}[/tex]
cosec(B) = [tex]\frac{25}{24}[/tex]
Since, 1 + cot²(B) = cosec²(B)
1 + cot²(B) = [tex](\frac{25}{24})^{2}[/tex]
cot²(B) = [tex]\frac{625}{576}-1[/tex]
cot(B) = [tex]\sqrt{\frac{49}{576}}[/tex]
= [tex]\pm \frac{7}{24}[/tex]
tan(B) = [tex]\pm \frac{24}{7}[/tex]
Since, tan and cot of an angle is negative in II quadrant,
Therefore, tan(B) = [tex]-\frac{24}{7}[/tex]
Since, tan(a + B) = [tex]\frac{\text{tan}(a)+\text{tan}(B)}{1-\text{tan(a)tan(B)}}[/tex]
By substituting the values in the identity,
tan(a + B) = [tex]\frac{-\frac{4}{3}-\frac{24}{7}}{1-(-\frac{4}{3})(-\frac{24}{7})}[/tex]
= [tex]\frac{-\frac{28}{21}-\frac{72}{21}}{1-(\frac{32}{7})}[/tex]
= [tex]\frac{-\frac{100}{21} }{\frac{7-32}{7} }[/tex]
= [tex]\frac{100}{21}\times \frac{7}{25}[/tex]
= [tex]\frac{4}{3}[/tex]
Therefore, tan(a + B) = [tex]\frac{4}{3}[/tex] is the answer.
