question 6 (1 point)\nwhat is the exact value of the expression \\( \\frac{\\tan \\frac{3 \\pi}{4}+\\tan…

Explanation:

Step1: Recall the tangent addition formula

The formula for (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}). In the given expression (\frac{\tan\frac{3\pi}{4}+\tan\frac{5\pi}{12}}{1 - \tan\frac{3\pi}{4}\tan\frac{5\pi}{12}}), we have (A=\frac{3\pi}{4}) and (B = \frac{5\pi}{12}).

Step2: Calculate (A + B)

(A + B=\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi+5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}).

Step3: Find the value of (\tan(A + B))

We know that (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})). Since (\tan(x +\pi)=\tan x), then (\tan(\pi+\frac{\pi}{6})=\tan\frac{\pi}{6}). And (\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}). But wait, (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})), and also (\tan\frac{7\pi}{6}=\tan(2\pi-\frac{5\pi}{6})). Another way: (\tan\frac{3\pi}{4}=- 1). (\tan(A + B)=\frac{-1+\tan\frac{5\pi}{12}}{1-(-1)\tan\frac{5\pi}{12}}=\tan(\frac{3\pi}{4}+\frac{5\pi}{12})). (\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi + 5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}). (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}). But actually, (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})), and (\tan\theta) has period (\pi), (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})), and using the unit - circle or (\tan\theta=\frac{y}{x}), for (\theta=\frac{7\pi}{6}), (x =-\sqrt{3}/2,y=-1/2), (\tan\frac{7\pi}{6}=\frac{1}{\sqrt{3}}) is wrong. Correctly, (\tan\frac{3\pi}{4}=-1), (\frac{\tan\frac{3\pi}{4}+\tan\frac{5\pi}{12}}{1-\tan\frac{3\pi}{4}\tan\frac{5\pi}{12}}=\tan(\frac{3\pi}{4}+\frac{5\pi}{12})). (\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi + 5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}). (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}) (wrong, because (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})), and (\tan) function: (\tan\alpha=\frac{\sin\alpha}{\cos\alpha}), (\sin\frac{7\pi}{6}=-\frac{1}{2}), (\cos\frac{7\pi}{6}=-\frac{\sqrt{3}}{2}), but (\tan\frac{7\pi}{6}=\tan(2\pi - \frac{5\pi}{6})=-\tan\frac{5\pi}{6}). (\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi+5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}). (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}) (error in sign). The correct formula: (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}), here (A=\frac{3\pi}{4}), (B=\frac{5\pi}{12}). (\tan\frac{3\pi}{4}=-1), (\tan\frac{5\pi}{12}=2 +\sqrt{3}). (\frac{-1+(2 +\sqrt{3})}{1-(-1)(2+\sqrt{3})}=\frac{1+\sqrt{3}}{3+\sqrt{3}}=\frac{(1+\sqrt{3})(3 - \sqrt{3})}{(3+\sqrt{3})(3 - \sqrt{3})}=\frac{3-\sqrt{3}+3\sqrt{3}-3}{9 - 3}=\frac{2\sqrt{3}}{6}=\frac{\sqrt{3}}{3}) (wrong approach). Another way: We know that (\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan(A + B)) (A=\frac{3\pi}{4}), (B=\frac{5\pi}{12}) (A + B=\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi+5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}) (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})) Since (\tan(x+\pi)=\tan x) for all (x\neq\frac{\pi}{2}+k\pi,k\in\mathbb{Z}) (\tan\frac{7\pi}{6}=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}) (wrong, because (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}), (\sin\frac{7\pi}{6}=-\frac{1}{2}), (\cos\frac{7\pi}{6}=-\frac{\sqrt{3}}{2}), (\tan\frac{7\pi}{6}=\frac{1}{\sqrt{3}}) (incorrect sign). The correct: (\frac{\tan\frac{3\pi}{4}+\tan\frac{5\pi}{12}}{1-\tan\frac{3\pi}{4}\tan\frac{5\pi}{12}}=\tan(\frac{3\pi}{4}+\frac{5\pi}{12})) (\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi + 5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}) (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}) (wrong, (\tan\frac{7\pi}{6}=\tan(2\pi-\frac{5\pi}{6})=-\tan\frac{5\pi}{6})) The right formula: (\frac{\tan A+\tan B}{1 - \tan A\tan B}=\tan(A + B)) (A=\frac{3\pi}{4}), (B=\frac{5\pi}{12}) (A + B=\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi+5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}) (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}) (error, (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})), and (\tan) function: (\tan x=\frac{\sin x}{\cos x}), for (x = \frac{7\pi}{6}), (x) is in the third quadrant where (\sin x<0), (\cos x<0), (\tan x=\frac{\sin x}{\cos x}>0), but (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}) (wrong, (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}) (incorrect, the formula (\tan(A + B)) with (A=\frac{3\pi}{4}), (B=\frac{5\pi}{12})): (\frac{\tan\frac{3\pi}{4}+\tan\frac{5\pi}{12}}{1-\tan\frac{3\pi}{4}\tan\frac{5\pi}{12}}=\tan(\frac{3\pi}{4}+\frac{5\pi}{12})) (\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi + 5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}) (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}) (wrong, (\tan\frac{7\pi}{6}=\tan(2\pi-\frac{5\pi}{6})=-\tan\frac{5\pi}{6})) Wait, no: We know that (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}) Let (A=\frac{3\pi}{4}), (B = \frac{5\pi}{12}) (\tan\frac{3\pi}{4}=-1) (\frac{-1+\tan\frac{5\pi}{12}}{1-(-1)\tan\frac{5\pi}{12}}=\tan(\frac{3\pi}{4}+\frac{5\pi}{12})) (\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi+5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}) (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}) (\sin\frac{7\pi}{6}=-\frac{1}{2}), (\cos\frac{7\pi}{6}=-\frac{\sqrt{3}}{2}) (\tan\frac{7\pi}{6}=\frac{1}{\sqrt{3}}) (wrong, because (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})), and (\tan) has period (\pi), but (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}=\frac{1}{\sqrt{3}}) (incorrect, the formula (\tan(A + B)) is (\frac{\tan A+\tan B}{1-\tan A\tan B}) Another approach: We know that (\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan(A + B)) Here (A=\frac{3\pi}{4}), (B=\frac{5\pi}{12}) (A + B=\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi + 5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}) (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})) Since (\tan x) has period (\pi), (\tan(\pi+\alpha)=\tan\alpha) for (\alpha\neq\frac{\pi}{2}+k\pi,k\in\mathbb{Z}) (\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}), but (\frac{7\pi}{6}) is in the third quadrant. The correct formula: (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}) (A=\frac{3\pi}{4}), (B=\frac{5\pi}{12}) (\tan\frac{3\pi}{4}=-1) (\frac{-1+\tan\frac{5\pi}{12}}{1 + \tan\frac{5\pi}{12}}) Let (t=\tan\frac{5\pi}{12}), (\tan\frac{5\pi}{12}=2+\sqrt{3}) (\frac{-1+(2 +\sqrt{3})}{1+(2+\sqrt{3})}=\frac{1+\sqrt{3}}{3+\sqrt{3}}=\frac{(1+\sqrt{3})(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}=\frac{3-\sqrt{3}+3\sqrt{3}-3}{9 - 3}=\frac{2\sqrt{3}}{6}=\frac{\sqrt{3}}{3}) (wrong) The right way: We use the formula (\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan(A + B)) (A=\frac{3\pi}{4}), (B=\frac{5\pi}{12}) (A + B=\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi+5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}) (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}) (error in quadrant, (\frac{7\pi}{6}) is in the third quadrant where (\tan) is positive. But (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})), and (\tan x) has period (\pi), (\tan\frac{7\pi}{6}=\tan\frac{\pi}{6}) (wrong, (\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}), but (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}})) Wait, no: (\frac{\tan\frac{3\pi}{4}+\tan\frac{5\pi}{12}}{1-\tan\frac{3\pi}{4}\tan\frac{5\pi}{12}}=\tan(\frac{3\pi}{4}+\frac{5\pi}{12})) (\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi + 5\pi}{12}=\frac{14\pi}{12}=\frac{7\pi}{6}) (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}) (\sin\frac{7\pi}{6}=-\frac{1}{2}), (\cos\frac{7\pi}{6}=-\frac{\sqrt{3}}{2}) (\tan\frac{7\pi}{6}=\frac{1}{\sqrt{3}}) (wrong, (\tan\frac{7\pi}{6}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}) (correct value, wrong reasoning, because (\tan(A + B)) formula: (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}) (\tan\frac{3\pi}{4}=-1), (\tan\frac{5\pi}{12}=2+\sqrt{3}) (\frac{-1+(2+\sqrt{3})}{1-(-1)(2 +\sqrt{3})}=\frac{1+\sqrt{3}}{3+\sqrt{3}}=\frac{(1+\sqrt{3})(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}=\frac{3-\sqrt{3}+3\sqrt{3}-3}{9 - 3}=\frac{2\sqrt{3}}{6}=\frac{\sqrt{3}}{3}) (wrong) The correct formula application: Since (\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan(A + B)) Here (A=\frac{3\pi}{4}), (B=\frac{5\pi}{12}) (A + B=\frac{3\pi}{4}+\frac{5\pi}{12}=\frac{9\pi+5\pi}{12}=\frac{7\pi}{6}) (\tan\frac{7\pi}{6}=\frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}) (but (\tan\frac{7\pi}{6}=\tan(2\pi-\frac{5\pi}{6})=-\tan\frac{