Find tan (a+B), given cot a= -3/4 , csc B= 25/24 , 90° < a < 180° , 90°< B < 180°

Question

Find tan (a+B), given cot a= -3/4 , csc B= 25/24 , 90° < a < 180° , 90°< B < 180°

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Tryphena 5 years 2021-07-28T18:50:51+00:00 1 Answers 13 views 0

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  1. MichaelMet
    0
    2021-07-28T18:52:13+00:00
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    Answer:

    tan(a + B) = \frac{4}{3}

    Step-by-step explanation:

    cot(a) = -\frac{3}{4}

    Therefore, tan(a) = -\frac{4}{3}

    cosec(B) = \frac{25}{24}

    Since, 1 + cot²(B) = cosec²(B)

    1 + cot²(B) = (\frac{25}{24})^{2}

    cot²(B) = \frac{625}{576}-1

    cot(B) = \sqrt{\frac{49}{576}}

              = \pm \frac{7}{24}

    tan(B) = \pm \frac{24}{7}

    Since, tan and cot of an angle is negative in II quadrant,

    Therefore, tan(B) = -\frac{24}{7}

    Since, tan(a + B) = \frac{\text{tan}(a)+\text{tan}(B)}{1-\text{tan(a)tan(B)}}

    By substituting the values in the identity,

    tan(a + B) = \frac{-\frac{4}{3}-\frac{24}{7}}{1-(-\frac{4}{3})(-\frac{24}{7})}

                    = \frac{-\frac{28}{21}-\frac{72}{21}}{1-(\frac{32}{7})}

                    = \frac{-\frac{100}{21} }{\frac{7-32}{7} }

                    = \frac{100}{21}\times \frac{7}{25}

                    = \frac{4}{3}

    Therefore, tan(a + B) = \frac{4}{3} is the answer.

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