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Question: If sin($\alpha + \beta$) = 1, sin($\alpha - \beta$) = $\frac{1}{2}$, then tan($\alpha + 2\beta$)tan(...

If sin(α+β\alpha + \beta) = 1, sin(α−β\alpha - \beta) = 12\frac{1}{2}, then tan(α+2β\alpha + 2\beta)tan(2α+β2\alpha + \beta) =

A

1

B

-1

C

zero

D

None of these

Answer

1

Explanation

Solution

To solve the problem, we first determine the values of (α+β)(\alpha + \beta) and (α−β)(\alpha - \beta) from the given sine equations.

Given:

  1. sin⁡(α+β)=1\sin(\alpha + \beta) = 1
  2. sin⁡(α−β)=12\sin(\alpha - \beta) = \frac{1}{2}

From equation (1), for the principal value, we have: α+β=π2\alpha + \beta = \frac{\pi}{2} (Equation A)

From equation (2), for the principal value, we have: α−β=π6\alpha - \beta = \frac{\pi}{6} (Equation B)

Now, we solve the system of linear equations (A) and (B) for α\alpha and β\beta.

Add Equation A and Equation B: (α+β)+(α−β)=π2+π6(\alpha + \beta) + (\alpha - \beta) = \frac{\pi}{2} + \frac{\pi}{6} 2α=3π+π62\alpha = \frac{3\pi + \pi}{6} 2α=4π62\alpha = \frac{4\pi}{6} 2α=2π32\alpha = \frac{2\pi}{3} α=π3\alpha = \frac{\pi}{3}

Substitute the value of α\alpha into Equation A: π3+β=π2\frac{\pi}{3} + \beta = \frac{\pi}{2} β=π2−π3\beta = \frac{\pi}{2} - \frac{\pi}{3} β=3π−2π6\beta = \frac{3\pi - 2\pi}{6} β=π6\beta = \frac{\pi}{6}

So, we have α=π3\alpha = \frac{\pi}{3} and β=π6\beta = \frac{\pi}{6}.

Next, we need to find the values of the arguments for the tangent functions: (α+2β)(\alpha + 2\beta) and (2α+β)(2\alpha + \beta).

Calculate (α+2β)(\alpha + 2\beta): α+2β=π3+2(π6)\alpha + 2\beta = \frac{\pi}{3} + 2\left(\frac{\pi}{6}\right) =π3+π3= \frac{\pi}{3} + \frac{\pi}{3} =2π3= \frac{2\pi}{3}

Calculate (2α+β)(2\alpha + \beta): 2α+β=2(π3)+π62\alpha + \beta = 2\left(\frac{\pi}{3}\right) + \frac{\pi}{6} =2π3+π6= \frac{2\pi}{3} + \frac{\pi}{6} =4π+π6= \frac{4\pi + \pi}{6} =5π6= \frac{5\pi}{6}

Now, calculate the tangent of these angles: tan⁡(α+2β)=tan⁡(2π3)\tan(\alpha + 2\beta) = \tan\left(\frac{2\pi}{3}\right) We know that tan⁡(π−x)=−tan⁡(x)\tan(\pi - x) = -\tan(x). tan⁡(2π3)=tan⁡(π−π3)=−tan⁡(π3)=−3\tan\left(\frac{2\pi}{3}\right) = \tan\left(\pi - \frac{\pi}{3}\right) = -\tan\left(\frac{\pi}{3}\right) = -\sqrt{3}

tan⁡(2α+β)=tan⁡(5π6)\tan(2\alpha + \beta) = \tan\left(\frac{5\pi}{6}\right) We know that tan⁡(π−x)=−tan⁡(x)\tan(\pi - x) = -\tan(x). tan⁡(5π6)=tan⁡(π−π6)=−tan⁡(π6)=−13\tan\left(\frac{5\pi}{6}\right) = \tan\left(\pi - \frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}

Finally, multiply the two tangent values: tan⁡(α+2β)tan⁡(2α+β)=(−3)×(−13)\tan(\alpha + 2\beta)\tan(2\alpha + \beta) = \left(-\sqrt{3}\right) \times \left(-\frac{1}{\sqrt{3}}\right) =1= 1