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Consider the following expression: (tan((25 pi)/(24))-tan((3pi)/(8)))/(1+tan((25 pi)/(24))tan((3pi)/(8))) Find the exact value of the expression without a calculator. [Hint: This diagram of special trigonometry values may help.]. Choose 1 answer: (A) -(sqrt3)/(3) (B) (sqrt3)/(3) (C) sqrt3 (D) -sqrt3

Consider the following expression:\newlinetan⁡(25π24)−tan⁡(3π8)1+tan⁡(25π24)tan⁡(3π8)\frac{\tan\left(\frac{25\pi}{24}\right)-\tan\left(\frac{3\pi}{8}\right)}{1+\tan\left(\frac{25\pi}{24}\right)\tan\left(\frac{3\pi}{8}\right)}\newlineFind the exact value of the expression without a calculator.\newline[Hint: This diagram of special trigonometry values may help.].\newlineChoose 11 answer:\newline(A)−33\text{(A)} \quad -\frac{\sqrt{3}}{3}\newline(B)33\text{(B)} \quad \frac{\sqrt{3}}{3}\newline(C)3\text{(C)} \quad \sqrt{3}\newline(D)−3\text{(D)} \quad -\sqrt{3}

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Full solution

Q. Consider the following expression:\newlinetan⁡(25π24)−tan⁡(3π8)1+tan⁡(25π24)tan⁡(3π8)\frac{\tan\left(\frac{25\pi}{24}\right)-\tan\left(\frac{3\pi}{8}\right)}{1+\tan\left(\frac{25\pi}{24}\right)\tan\left(\frac{3\pi}{8}\right)}\newlineFind the exact value of the expression without a calculator.\newline[Hint: This diagram of special trigonometry values may help.].\newlineChoose 11 answer:\newline(A)−33\text{(A)} \quad -\frac{\sqrt{3}}{3}\newline(B)33\text{(B)} \quad \frac{\sqrt{3}}{3}\newline(C)3\text{(C)} \quad \sqrt{3}\newline(D)−3\text{(D)} \quad -\sqrt{3}
  1. Tangent Subtraction Formula: We will use the tangent subtraction formula, which is given by:\newlinetan⁡(A−B)=tan⁡(A)−tan⁡(B)1+tan⁡(A)tan⁡(B)\tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)}\newlineHere, A=25π24A = \frac{25 \pi}{24} and B=3π8B = \frac{3 \pi}{8}.
  2. Simplify Angles: First, we need to simplify the angles (25π)/24(25 \pi)/24 and (3π)/8(3 \pi)/8 to find their tangent values. We can do this by expressing them in terms of known special angles.\newline(25π)/24=(24π)/24+(π)/24=π+(π)/24(25 \pi)/24 = (24 \pi)/24 + (\pi)/24 = \pi + (\pi)/24\newline(3π)/8=(2π)/8+(π)/8=(π)/4+(π)/8(3 \pi)/8 = (2 \pi)/8 + (\pi)/8 = (\pi)/4 + (\pi)/8
  3. Find Tangent Values: Now, we need to find the tangent values for these angles. Since π+(π)/24\pi + (\pi)/24 is more than π\pi, we can use the periodicity of the tangent function to find that tan⁡(π+(π)/24)=tan⁡((π)/24)\tan(\pi + (\pi)/24) = \tan((\pi)/24). Similarly, tan⁡((π)/4+(π)/8)=tan⁡((3π)/8)\tan((\pi)/4 + (\pi)/8) = \tan((3\pi)/8).
  4. Find Tangent of (π)/24(\pi)/24: The tangent of (π)/24(\pi)/24 is not a standard angle, but we can use the fact that (π)/24(\pi)/24 is half of (π)/12(\pi)/12, which corresponds to 1515 degrees, and tan⁡(15\tan(15 degrees) is a known value. However, we do not have a direct value for tan⁡((π)/24)\tan((\pi)/24), so we need to use the tangent addition formula to find tan⁡((π)/12)\tan((\pi)/12) first and then use the half-angle formula.
  5. Quadratic Equation for x: Using the tangent addition formula, we have:\newlinetan⁡(π12)=tan⁡(π6+π12)=tan⁡(π6)+tan⁡(π12)1−tan⁡(π6)tan⁡(π12)\tan(\frac{\pi}{12}) = \tan(\frac{\pi}{6} + \frac{\pi}{12}) = \frac{\tan(\frac{\pi}{6}) + \tan(\frac{\pi}{12})}{1 - \tan(\frac{\pi}{6})\tan(\frac{\pi}{12})}\newlineWe know that tan⁡(π6)=13\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}. Let's denote tan⁡(π12)\tan(\frac{\pi}{12}) as xx and solve for xx.
  6. Solve Quadratic Equation: x=1/3+x1−(1/3)xx = \frac{1/\sqrt{3} + x}{1 - (1/\sqrt{3})x}x−x3=13+x23x - \frac{x}{\sqrt{3}} = \frac{1}{\sqrt{3}} + \frac{x^2}{\sqrt{3}}x(1−13)=13+x23x(1 - \frac{1}{\sqrt{3}}) = \frac{1}{\sqrt{3}} + \frac{x^2}{\sqrt{3}}x(3−1)=1+x2x(\sqrt{3} - 1) = 1 + x^2x2+x(3−1)−1=0x^2 + x(\sqrt{3} - 1) - 1 = 0This is a quadratic equation in xx.
  7. Positive Solution for (π)/12(\pi)/12: We can solve the quadratic equation using the quadratic formula:\newlinex=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\newlineHere, a=1a = 1, b=3−1b = \sqrt{3} - 1, and c=−1c = -1.
  8. Find Tangent of (π)/24(\pi)/24: Plugging the values into the quadratic formula, we get:\newlinex=−(3−1)±(3−1)2−4(1)(−1)2(1)x = \frac{-\left(\sqrt{3} - 1\right) \pm \sqrt{\left(\sqrt{3} - 1\right)^2 - 4(1)(-1)}}{2(1)}\newlinex=−(3−1)±3−23+1+42x = \frac{-\left(\sqrt{3} - 1\right) \pm \sqrt{3 - 2\sqrt{3} + 1 + 4}}{2}\newlinex=−(3−1)±8−232x = \frac{-\left(\sqrt{3} - 1\right) \pm \sqrt{8 - 2\sqrt{3}}}{2}
  9. Cosine Addition Formula: We need the positive solution for tan⁡(π12)\tan\left(\frac{\pi}{12}\right) because π12\frac{\pi}{12} is in the first quadrant where tangent is positive. So we choose the plus sign in the quadratic formula.\newlinex=−(3−1)+8−232x = \frac{-\left(\sqrt{3} - 1\right) + \sqrt{8 - 2\sqrt{3}}}{2}
  10. Find sin⁡(π12):</b>Nowweneedtofind$tan⁡(π24)\sin\left(\frac{\pi}{12}\right):</b> Now we need to find \$\tan\left(\frac{\pi}{24}\right) using the half-angle formula:\newlinetan⁡(π24)=(1−cos⁡(π12)1+cos⁡(π12))\tan\left(\frac{\pi}{24}\right) = \sqrt{\left(\frac{1 - \cos\left(\frac{\pi}{12}\right)}{1 + \cos\left(\frac{\pi}{12}\right)}\right)}\newlineWe know that cos⁡(π12)\cos\left(\frac{\pi}{12}\right) can be found using the cosine addition formula for cos⁡(π6−π12)\cos\left(\frac{\pi}{6} - \frac{\pi}{12}\right).
  11. Solve Equation for y: Using the cosine addition formula, we have:\newlinecos⁡(π12)=cos⁡(π6−π12)=cos⁡(π6)cos⁡(π12)+sin⁡(π6)sin⁡(π12)\cos\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{6} - \frac{\pi}{12}\right) = \cos\left(\frac{\pi}{6}\right)\cos\left(\frac{\pi}{12}\right) + \sin\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{12}\right)\newlineWe know that cos⁡(π6)=32\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} and sin⁡(π6)=12\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}. Let's denote cos⁡(π12)\cos\left(\frac{\pi}{12}\right) as yy and solve for yy.
  12. Solve Equation for y: Using the cosine addition formula, we have:\newlinecos⁡(π12)=cos⁡(π6−π12)=cos⁡(π6)cos⁡(π12)+sin⁡(π6)sin⁡(π12)\cos(\frac{\pi}{12}) = \cos(\frac{\pi}{6} - \frac{\pi}{12}) = \cos(\frac{\pi}{6})\cos(\frac{\pi}{12}) + \sin(\frac{\pi}{6})\sin(\frac{\pi}{12})\newlineWe know that cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} and sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2}. Let's denote cos⁡(π12)\cos(\frac{\pi}{12}) as yy and solve for yy.y=(32)y+(12)sin⁡(π12)y = (\frac{\sqrt{3}}{2})y + (\frac{1}{2})\sin(\frac{\pi}{12})\newlineTo find sin⁡(π12)\sin(\frac{\pi}{12}), we can use the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1.\newlinesin⁡2(π12)=1−cos⁡2(π12)\sin^2(\frac{\pi}{12}) = 1 - \cos^2(\frac{\pi}{12})\newlinecos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}00
  13. Solve Equation for y: Using the cosine addition formula, we have:\newlinecos⁡(π12)=cos⁡(π6−π12)=cos⁡(π6)cos⁡(π12)+sin⁡(π6)sin⁡(π12)\cos(\frac{\pi}{12}) = \cos(\frac{\pi}{6} - \frac{\pi}{12}) = \cos(\frac{\pi}{6})\cos(\frac{\pi}{12}) + \sin(\frac{\pi}{6})\sin(\frac{\pi}{12})\newlineWe know that cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} and sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2}. Let's denote cos⁡(π12)\cos(\frac{\pi}{12}) as yy and solve for yy.y=(32)y+(12)sin⁡(π12)y = (\frac{\sqrt{3}}{2})y + (\frac{1}{2})\sin(\frac{\pi}{12})\newlineTo find sin⁡(π12)\sin(\frac{\pi}{12}), we can use the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1.\newlinesin⁡2(π12)=1−cos⁡2(π12)\sin^2(\frac{\pi}{12}) = 1 - \cos^2(\frac{\pi}{12})\newlinecos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}00Substituting back into the equation for yy, we get:\newlinecos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}22\newlinecos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}33\newlinecos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}44\newlineThis is another equation we need to solve for yy.
  14. Solve Equation for y: Using the cosine addition formula, we have:\newlinecos⁡(π12)=cos⁡(π6−π12)=cos⁡(π6)cos⁡(π12)+sin⁡(π6)sin⁡(π12)\cos(\frac{\pi}{12}) = \cos(\frac{\pi}{6} - \frac{\pi}{12}) = \cos(\frac{\pi}{6})\cos(\frac{\pi}{12}) + \sin(\frac{\pi}{6})\sin(\frac{\pi}{12})\newlineWe know that cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} and sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2}. Let's denote cos⁡(π12)\cos(\frac{\pi}{12}) as yy and solve for yy.y=(32)y+(12)sin⁡(π12)y = (\frac{\sqrt{3}}{2})y + (\frac{1}{2})\sin(\frac{\pi}{12})\newlineTo find sin⁡(π12)\sin(\frac{\pi}{12}), we can use the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1.\newlinesin⁡2(π12)=1−cos⁡2(π12)\sin^2(\frac{\pi}{12}) = 1 - \cos^2(\frac{\pi}{12})\newlinecos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}00Substituting back into the equation for yy, we get:\newlinecos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}22\newlinecos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}33\newlinecos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}44\newlineThis is another equation we need to solve for yy.We made a mistake in the previous steps. The tangent of cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}66 is not a standard angle, and we cannot find its exact value using the half-angle formula from cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}77 because we do not have an exact value for cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}77. We need to reconsider our approach to find the tangent values for cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}99 and sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2}00.

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