use the given information to find (a) \\( \\sin (s+t) \\), (b) \\( \\tan (s+t) \\), and (c) the quadrant of…

use the given information to find (a) \\( \\sin (s+t) \\), (b) \\( \\tan (s+t) \\), and (c) the quadrant of \\( s+t \\).\n\\( \\cos s=\\frac{12}{13} \\) and \\( \\sin t=\\frac{3}{5} \\), s and t in quadrant i\n(a) \\( \\sin (s+t)=\\frac{56}{65} \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)\n(b) \\( \\tan (s+t)=\\frac{56}{33} \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)\n(c) what is the quadrant of \\( s+t \\)?\nquadrant iv\nquadrant i\nquadrant ii\nquadrant iii

use the given information to find (a) \\( \\sin (s+t) \\), (b) \\( \\tan (s+t) \\), and (c) the quadrant of \\( s+t \\).\n\\( \\cos s=\\frac{12}{13} \\) and \\( \\sin t=\\frac{3}{5} \\), s and t in quadrant i\n(a) \\( \\sin (s+t)=\\frac{56}{65} \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)\n(b) \\( \\tan (s+t)=\\frac{56}{33} \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)\n(c) what is the quadrant of \\( s+t \\)?\nquadrant iv\nquadrant i\nquadrant ii\nquadrant iii

Answer

Explanation:

Step1: Determine (\sin s) and (\cos t)

Since (s) and (t) are in quadrant I, (\sin s=\sqrt{1 - \cos^{2}s}=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{169 - 144}{169}}=\frac{5}{13}) (\cos t=\sqrt{1-\sin^{2}t}=\sqrt{1 - \left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{25 - 9}{25}}=\frac{4}{5})

Step2: Calculate (\tan s) and (\tan t)

(\tan s=\frac{\sin s}{\cos s}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}) (\tan t=\frac{\sin t}{\cos t}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4})

Step3: Analyze the quadrant of (s + t)

We know that (\sin(s + t)=\frac{56}{65}>0) and (\tan(s + t)=\frac{56}{33}>0) In the unit - circle, if (\sin\alpha>0) and (\tan\alpha>0), then (\cos\alpha>0) (because (\tan\alpha=\frac{\sin\alpha}{\cos\alpha})). In quadrant I, (\sin\theta>0), (\cos\theta>0) and (\tan\theta>0); in quadrant II, (\sin\theta>0), (\cos\theta<0) and (\tan\theta<0); in quadrant III, (\sin\theta<0), (\cos\theta<0) and (\tan\theta>0); in quadrant IV, (\sin\theta<0), (\cos\theta>0) and (\tan\theta<0)

Answer:

(c) Quadrant I