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Question: If tan²$\alpha$ + 2tan$\alpha$.tan2$\beta$ = tan²$\beta$ + 2tan$\beta$.tan2$\alpha$, then we may hav...

If tan²α\alpha + 2tanα\alpha.tan2β\beta = tan²β\beta + 2tanβ\beta.tan2α\alpha, then we may have

A

tan²α\alpha + 2tanα\alpha.tan2β\beta = 0

B

tanα\alpha + tanβ\beta = 0

C

tan²β\beta + 2tanβ\beta.tan2α\alpha = 1

D

tanα\alpha = tanβ\beta

Answer

B, D

Explanation

Solution

The given equation is: tan⁡2α+2tan⁡αtan⁡2β=tan⁡2β+2tan⁡βtan⁡2α\tan^2\alpha + 2\tan\alpha\tan2\beta = \tan^2\beta + 2\tan\beta\tan2\alpha

Rearrange the terms: tan⁡2α−tan⁡2β=2tan⁡βtan⁡2α−2tan⁡αtan⁡2β\tan^2\alpha - \tan^2\beta = 2\tan\beta\tan2\alpha - 2\tan\alpha\tan2\beta (tan⁡α−tan⁡β)(tan⁡α+tan⁡β)=−2(tan⁡αtan⁡2β−tan⁡βtan⁡2α)(\tan\alpha - \tan\beta)(\tan\alpha + \tan\beta) = -2(\tan\alpha\tan2\beta - \tan\beta\tan2\alpha)

Let a=tan⁡αa = \tan\alpha and b=tan⁡βb = \tan\beta. The equation becomes: (a−b)(a+b)=−2(atan⁡2β−btan⁡2α)(a-b)(a+b) = -2(a\tan2\beta - b\tan2\alpha)

Now, substitute the double angle formulas: tan⁡2x=2tan⁡x1−tan⁡2x\tan2x = \frac{2\tan x}{1-\tan^2 x}. So, tan⁡2α=2a1−a2\tan2\alpha = \frac{2a}{1-a^2} and tan⁡2β=2b1−b2\tan2\beta = \frac{2b}{1-b^2}. The equation becomes: (a−b)(a+b)=−2(a2b1−b2−b2a1−a2)(a-b)(a+b) = -2\left(a\frac{2b}{1-b^2} - b\frac{2a}{1-a^2}\right) (a−b)(a+b)=−2(2ab1−b2−2ab1−a2)(a-b)(a+b) = -2\left(\frac{2ab}{1-b^2} - \frac{2ab}{1-a^2}\right) (a−b)(a+b)=−4ab(11−b2−11−a2)(a-b)(a+b) = -4ab\left(\frac{1}{1-b^2} - \frac{1}{1-a^2}\right) (a−b)(a+b)=−4ab((1−a2)−(1−b2)(1−b2)(1−a2))(a-b)(a+b) = -4ab\left(\frac{(1-a^2) - (1-b^2)}{(1-b^2)(1-a^2)}\right) (a−b)(a+b)=−4ab(b2−a2(1−a2)(1−b2))(a-b)(a+b) = -4ab\left(\frac{b^2 - a^2}{(1-a^2)(1-b^2)}\right) (a−b)(a+b)=−4ab(−(a2−b2)(1−a2)(1−b2))(a-b)(a+b) = -4ab\left(\frac{-(a^2 - b^2)}{(1-a^2)(1-b^2)}\right) (a−b)(a+b)=4ab((a−b)(a+b)(1−a2)(1−b2))(a-b)(a+b) = 4ab\left(\frac{(a-b)(a+b)}{(1-a^2)(1-b^2)}\right)

Now, we have two possibilities:

Case 1: (a−b)(a+b)=0(a-b)(a+b) = 0 This implies a−b=0a-b=0 or a+b=0a+b=0. If a−b=0a-b=0, then a=ba=b, which means tan⁡α=tan⁡β\tan\alpha = \tan\beta. This matches option (D). If a+b=0a+b=0, then a=−ba=-b, which means tan⁡α=−tan⁡β\tan\alpha = -\tan\beta, or tan⁡α+tan⁡β=0\tan\alpha + \tan\beta = 0. This matches option (B).

Case 2: (a−b)(a+b)≠0(a-b)(a+b) \neq 0 In this case, we can divide both sides by (a−b)(a+b)(a-b)(a+b): 1=4ab(1−a2)(1−b2)1 = \frac{4ab}{(1-a^2)(1-b^2)} (1−a2)(1−b2)=4ab(1-a^2)(1-b^2) = 4ab 1−a2−b2+a2b2=4ab1 - a^2 - b^2 + a^2b^2 = 4ab 1=a2+b2−a2b2+4ab1 = a^2 + b^2 - a^2b^2 + 4ab

Let's check if this third case leads to any of the options. If we substitute tan⁡α=tan⁡β\tan\alpha = \tan\beta (from Case 1) into this condition, we get: (1−a2)2=4a2(1-a^2)^2 = 4a^2 1−a2=±2a1-a^2 = \pm 2a a2±2a−1=0a^2 \pm 2a - 1 = 0 For a2+2a−1=0a^2+2a-1=0, a=−2±4−4(1)(−1)2=−1±2a = \frac{-2 \pm \sqrt{4-4(1)(-1)}}{2} = -1 \pm \sqrt{2}. For a2−2a−1=0a^2-2a-1=0, a=2±4−4(1)(−1)2=1±2a = \frac{2 \pm \sqrt{4-4(1)(-1)}}{2} = 1 \pm \sqrt{2}. So, if tan⁡α=tan⁡β=±1±2\tan\alpha = \tan\beta = \pm 1 \pm \sqrt{2}, these are specific solutions that also satisfy tan⁡α=tan⁡β\tan\alpha=\tan\beta. These values are valid since tan⁡α≠±1\tan\alpha \neq \pm 1.

If we substitute tan⁡α=−tan⁡β\tan\alpha = -\tan\beta (from Case 1) into this condition, we get: Let b=−ab=-a. (1−a2)(1−(−a)2)=4a(−a)(1-a^2)(1-(-a)^2) = 4a(-a) (1−a2)2=−4a2(1-a^2)^2 = -4a^2 (1−a2)2+4a2=0(1-a^2)^2 + 4a^2 = 0 Since (1−a2)2≥0(1-a^2)^2 \ge 0 and 4a2≥04a^2 \ge 0, their sum can only be zero if both terms are zero. This implies a2=0a^2=0 and (1−a2)2=0(1-a^2)^2=0. a=0a=0 from a2=0a^2=0. Substituting a=0a=0 into (1−a2)2=0(1-a^2)^2=0 gives (1−0)2=0(1-0)^2=0, which means 1=01=0. This is a contradiction. The only way for (1−a2)2+4a2=0(1-a^2)^2 + 4a^2 = 0 to hold is if aa is imaginary, which is not possible for real α\alpha. However, if a=0a=0, then b=0b=0. In this case, tan⁡α=0\tan\alpha=0 and tan⁡β=0\tan\beta=0. This satisfies both tan⁡α=tan⁡β\tan\alpha=\tan\beta and tan⁡α+tan⁡β=0\tan\alpha+\tan\beta=0.

So, the general solutions are tan⁡α=tan⁡β\tan\alpha = \tan\beta or tan⁡α+tan⁡β=0\tan\alpha + \tan\beta = 0. Both options (B) and (D) are derived from the given equation. Since the question asks "then we may have", any of these valid conditions is a correct answer.