Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Answer:
[tex] \tan (A +B) = - 6\frac{6}{13} [/tex]
Step-by-step explanation:
[tex] \sin \: A = \frac{3}{5}...(given) \\ \\ \because\cos A = \pm \sqrt{1 - { \sin }^{2}\: A } \\ \\ \ \therefore\cos A = \pm \sqrt{1 - \bigg( \frac{3}{5} \bigg) ^{2}} \\ \\ \ \therefore\cos A = \pm \sqrt{1 - \frac{9}{25}} \\ \\ \therefore\cos A = \pm \sqrt{ \frac{25 - 9}{25} } \\ \\ \therefore\cos A = \pm \sqrt{ \frac{16}{25}} \\ \\ \therefore\cos A = \pm { \frac{4}{5}} \\ \\ \because \: \angle A \: is \: in \: quadrant \: I \\ \\ \therefore \: \cos A = { \frac{4}{5}} \\ \\ \tan \: A = \frac{ \sin \: A}{\cos \: A} \\ \\ \tan \: A = \frac{ \frac{3}{5} }{ \frac{4}{5} } \\ \\ \tan \: A = \frac{3}{4} \\ \\ \because \tan (A +B) = \frac{ \tan A + \tan B }{1 - \tan A \tan B } \\ \\ \therefore \tan (A +B) = \frac{ \frac{3}{4} + \frac{15}{8} }{1 - \frac{3}{4} \times \frac{15}{8} } \\ \\ \therefore \tan (A +B) = \frac{ \frac{6}{8} + \frac{15}{8} }{1 - \frac{45}{32}} \\ \\ \therefore \tan (A +B) = \frac{ \frac{21}{8} }{\frac{32 - 45}{32}} \\ \\ \therefore \tan (A +B) = \frac{ \frac{21}{8} }{\frac{ - 13}{32}} \\ \\ \therefore \tan (A +B) = \frac{21}{8} \times \bigg(- \frac{32}{13 } \bigg)\\ \\ \therefore \tan (A +B) = - \frac{84}{13} \\ \\ \therefore \tan (A +B) = - 6\frac{6}{13} [/tex]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
