Answer:
See answers below
Step-by-step explanation:
Problem 1
Recall that [tex]tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}[/tex] and that [tex]tan(\frac{\pi}{4})=1[/tex]. Using these two facts, we can rewrite the expression:
[tex]\frac{1+tanx}{-1+tanx}\\\\-\frac{1+tanx}{1-tanx}\\ \\-\frac{tan(\frac{\pi}{4})+tanx}{1-tan(\frac{\pi}{4})tan(x)}\\ \\-tan(x+\frac{\pi}{4})\\ \\tan(\frac{3\pi}{4}-x)[/tex]
Hence, the first choice is correct
Problem 2
[tex]cos(\frac{x}{2})=\sqrt{3}-cos(\frac{x}{2})\:;\: 0\leq x < 360^\circ\\\\2cos(\frac{x}{2})=\sqrt{3}\\ \\cos(\frac{x}{2})=\frac{\sqrt{3}}{2}\\\\\frac{x}{2}=\frac{\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n\\ \\ x=\frac{\pi}{3}+4\pi n,\frac{11\pi}{3}+4\pi n\\ \\ x=60^\circ+720n^\circ, 660^\circ+720n^\circ\\\\x=60^\circ[/tex]
It's helpful to use the unit circle to solve these kinds of problems. Therefore, the third answer is correct.
Problem 3
Because [tex]sin\theta=-\frac{5}{13}[/tex] and our parameters are [tex]\pi < \theta < \frac{3\pi}{2}[/tex], the triangle must be in Quadrant III where [tex]sin\theta < 0[/tex] and [tex]cos\theta < 0[/tex].
You may recall the double angle formula [tex]sin2\theta=2sin\theta cos\theta[/tex]. We can find [tex]cos\theta[/tex] using [tex]sin\theta[/tex] with the Pythagorean Identity [tex]sin^2\theta+cos^2\theta=1[/tex] keeping our parameters in mind:
[tex]sin^2\theta+cos^2\theta=1\\\\(-\frac{5}{13})^2+cos^2\theta=1\\ \\\frac{25}{169}+cos^2\theta=1\\ \\cos^2\theta=\frac{144}{169}\\ \\cos\theta=-\frac{12}{13}[/tex]
Thus, [tex]sin2\theta=2sin\theta cos\theta=2(-\frac{5}{13})(-\frac{12}{13})=2(\frac{60}{169})=\frac{120}{169}[/tex], which means the third option is correct.
