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Question: If $\frac{x}{a}cosa + \frac{y}{b}sin \alpha = 1, \frac{x}{a}cos\beta + \frac{y}{b}sin \beta = 1$ and...

If xacosa+ybsinα=1,xacosβ+ybsinβ=1\frac{x}{a}cosa + \frac{y}{b}sin \alpha = 1, \frac{x}{a}cos\beta + \frac{y}{b}sin \beta = 1 and cosαcosβa2+sinαsinβb2=0\frac{cos\alpha cos\beta}{a^2} + \frac{sin \alpha sin \beta}{b^2} = 0; α−β≠2nπ,n∈Z\alpha - \beta \neq 2n\pi, n \in Z then prove that -

tanαtanβ=b2(x2−a2)a2(y2−b2)tan \alpha tan \beta = \frac{b^2(x^2-a^2)}{a^2(y^2-b^2)} and x2+y2=a2+b2x^2 + y^2 = a^2 + b^2

Answer

The problem asks us to prove two statements given three conditions.

Given Conditions:

  1. xacos⁡α+ybsin⁡α=1\frac{x}{a}\cos\alpha + \frac{y}{b}\sin \alpha = 1
  2. xacos⁡β+ybsin⁡β=1\frac{x}{a}\cos\beta + \frac{y}{b}\sin \beta = 1
  3. cos⁡αcos⁡βa2+sin⁡αsin⁡βb2=0\frac{\cos\alpha \cos\beta}{a^2} + \frac{\sin \alpha \sin \beta}{b^2} = 0
  4. α−β≠2nπ,n∈Z\alpha - \beta \neq 2n\pi, n \in Z

Statements to Prove: A. tan⁡αtan⁡β=b2(x2−a2)a2(y2−b2)\tan \alpha \tan \beta = \frac{b^2(x^2-a^2)}{a^2(y^2-b^2)} B. x2+y2=a2+b2x^2 + y^2 = a^2 + b^2

Step-by-step Derivation:

Part 1: Solve for x/a and y/b Let X=xaX = \frac{x}{a} and Y=ybY = \frac{y}{b}. The first two equations become: Xcos⁡α+Ysin⁡α=1(i)X \cos\alpha + Y \sin \alpha = 1 \quad (i) Xcos⁡β+Ysin⁡β=1(ii)X \cos\beta + Y \sin \beta = 1 \quad (ii)

Subtracting (ii) from (i): X(cos⁡α−cos⁡β)+Y(sin⁡α−sin⁡β)=0X(\cos\alpha - \cos\beta) + Y(\sin \alpha - \sin \beta) = 0 Using sum-to-product formulas: X(−2sin⁡α+β2sin⁡α−β2)+Y(2cos⁡α+β2sin⁡α−β2)=0X(-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}) + Y(2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}) = 0 Since α−β≠2nπ\alpha - \beta \neq 2n\pi, sin⁡α−β2≠0\sin\frac{\alpha-\beta}{2} \neq 0. We can divide by 2sin⁡α−β22\sin\frac{\alpha-\beta}{2}: −Xsin⁡α+β2+Ycos⁡α+β2=0-X\sin\frac{\alpha+\beta}{2} + Y\cos\frac{\alpha+\beta}{2} = 0 Ycos⁡α+β2=Xsin⁡α+β2Y\cos\frac{\alpha+\beta}{2} = X\sin\frac{\alpha+\beta}{2} YX=tan⁡α+β2  ⟹  aybx=tan⁡α+β2\frac{Y}{X} = \tan\frac{\alpha+\beta}{2} \implies \frac{ay}{bx} = \tan\frac{\alpha+\beta}{2}

Alternatively, we can solve for X and Y using Cramer's Rule or substitution. Multiply (i) by sin⁡β\sin\beta and (ii) by sin⁡α\sin\alpha: Xcos⁡αsin⁡β+Ysin⁡αsin⁡β=sin⁡βX \cos\alpha \sin\beta + Y \sin \alpha \sin\beta = \sin\beta Xcos⁡βsin⁡α+Ysin⁡βsin⁡α=sin⁡αX \cos\beta \sin\alpha + Y \sin \beta \sin\alpha = \sin\alpha Subtracting the second from the first: X(cos⁡αsin⁡β−cos⁡βsin⁡α)=sin⁡β−sin⁡αX(\cos\alpha \sin\beta - \cos\beta \sin\alpha) = \sin\beta - \sin\alpha Xsin⁡(β−α)=sin⁡β−sin⁡αX \sin(\beta - \alpha) = \sin\beta - \sin\alpha X=sin⁡β−sin⁡αsin⁡(β−α)=2cos⁡α+β2sin⁡β−α22sin⁡β−α2cos⁡β−α2=cos⁡α+β2cos⁡α−β2X = \frac{\sin\beta - \sin\alpha}{\sin(\beta - \alpha)} = \frac{2\cos\frac{\alpha+\beta}{2}\sin\frac{\beta-\alpha}{2}}{2\sin\frac{\beta-\alpha}{2}\cos\frac{\beta-\alpha}{2}} = \frac{\cos\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}} So, xa=cos⁡α+β2cos⁡α−β2(v)\frac{x}{a} = \frac{\cos\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}} \quad (v)

Multiply (i) by cos⁡β\cos\beta and (ii) by cos⁡α\cos\alpha: Xcos⁡αcos⁡β+Ysin⁡αcos⁡β=cos⁡βX \cos\alpha \cos\beta + Y \sin \alpha \cos\beta = \cos\beta Xcos⁡βcos⁡α+Ysin⁡βcos⁡α=cos⁡αX \cos\beta \cos\alpha + Y \sin \beta \cos\alpha = \cos\alpha Subtracting the first from the second: Y(sin⁡βcos⁡α−sin⁡αcos⁡β)=cos⁡α−cos⁡βY(\sin \beta \cos\alpha - \sin \alpha \cos\beta) = \cos\alpha - \cos\beta Ysin⁡(β−α)=cos⁡α−cos⁡βY \sin(\beta - \alpha) = \cos\alpha - \cos\beta Y=cos⁡α−cos⁡βsin⁡(β−α)=−2sin⁡α+β2sin⁡α−β2−2sin⁡α−β2cos⁡α−β2=sin⁡α+β2cos⁡α−β2Y = \frac{\cos\alpha - \cos\beta}{\sin(\beta - \alpha)} = \frac{-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}}{-2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha-\beta}{2}} = \frac{\sin\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}} So, yb=sin⁡α+β2cos⁡α−β2(vi)\frac{y}{b} = \frac{\sin\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}} \quad (vi) Note: Since α−β≠2nπ\alpha - \beta \neq 2n\pi, cos⁡α−β2≠0\cos\frac{\alpha-\beta}{2} \neq 0. If cos⁡α−β2=0\cos\frac{\alpha-\beta}{2} = 0, then xx and yy would be undefined.

Part 2: Prove x2+y2=a2+b2x^2+y^2 = a^2+b^2 Square equations (v) and (vi) and add them: (xa)2+(yb)2=cos⁡2α+β2cos⁡2α−β2+sin⁡2α+β2cos⁡2α−β2\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = \frac{\cos^2\frac{\alpha+\beta}{2}}{\cos^2\frac{\alpha-\beta}{2}} + \frac{\sin^2\frac{\alpha+\beta}{2}}{\cos^2\frac{\alpha-\beta}{2}} x2a2+y2b2=cos⁡2α+β2+sin⁡2α+β2cos⁡2α−β2=1cos⁡2α−β2\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{\cos^2\frac{\alpha+\beta}{2} + \sin^2\frac{\alpha+\beta}{2}}{\cos^2\frac{\alpha-\beta}{2}} = \frac{1}{\cos^2\frac{\alpha-\beta}{2}} x2a2+y2b2=11+cos⁡(α−β)2=21+cos⁡(α−β)(vii)\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{1}{\frac{1+\cos(\alpha-\beta)}{2}} = \frac{2}{1+\cos(\alpha-\beta)} \quad (vii)

Now, let's use the third given condition: cos⁡αcos⁡βa2+sin⁡αsin⁡βb2=0\frac{\cos\alpha \cos\beta}{a^2} + \frac{\sin \alpha \sin \beta}{b^2} = 0 Multiply by a2b2a^2 b^2: b2cos⁡αcos⁡β+a2sin⁡αsin⁡β=0b^2 \cos\alpha \cos\beta + a^2 \sin \alpha \sin \beta = 0 Divide by cos⁡αcos⁡β\cos\alpha \cos\beta (assuming cos⁡αcos⁡β≠0\cos\alpha \cos\beta \neq 0, if it were zero, then sin⁡αsin⁡β\sin\alpha \sin\beta would not be zero and the equality would not hold unless a2=0a^2=0 or b2=0b^2=0, which is not the case): b2+a2sin⁡αsin⁡βcos⁡αcos⁡β=0b^2 + a^2 \frac{\sin \alpha \sin \beta}{\cos\alpha \cos\beta} = 0 b2+a2tan⁡αtan⁡β=0b^2 + a^2 \tan \alpha \tan \beta = 0 tan⁡αtan⁡β=−b2a2(viii)\tan \alpha \tan \beta = -\frac{b^2}{a^2} \quad (viii)

We know the identities: cos⁡αcos⁡β=12[cos⁡(α−β)+cos⁡(α+β)]\cos\alpha \cos\beta = \frac{1}{2}[\cos(\alpha-\beta) + \cos(\alpha+\beta)] sin⁡αsin⁡β=12[cos⁡(α−β)−cos⁡(α+β)]\sin \alpha \sin \beta = \frac{1}{2}[\cos(\alpha-\beta) - \cos(\alpha+\beta)] Substitute these into the third given condition: 12[cos⁡(α−β)+cos⁡(α+β)]a2+12[cos⁡(α−β)−cos⁡(α+β)]b2=0\frac{\frac{1}{2}[\cos(\alpha-\beta) + \cos(\alpha+\beta)]}{a^2} + \frac{\frac{1}{2}[\cos(\alpha-\beta) - \cos(\alpha+\beta)]}{b^2} = 0 Multiply by 2a2b22a^2b^2: b2[cos⁡(α−β)+cos⁡(α+β)]+a2[cos⁡(α−β)−cos⁡(α+β)]=0b^2[\cos(\alpha-\beta) + \cos(\alpha+\beta)] + a^2[\cos(\alpha-\beta) - \cos(\alpha+\beta)] = 0 (b2+a2)cos⁡(α−β)+(b2−a2)cos⁡(α+β)=0(b^2+a^2)\cos(\alpha-\beta) + (b^2-a^2)\cos(\alpha+\beta) = 0 (a2+b2)cos⁡(α−β)=(a2−b2)cos⁡(α+β)(a^2+b^2)\cos(\alpha-\beta) = (a^2-b^2)\cos(\alpha+\beta) cos⁡(α+β)=a2+b2a2−b2cos⁡(α−β)(ix)\cos(\alpha+\beta) = \frac{a^2+b^2}{a^2-b^2}\cos(\alpha-\beta) \quad (ix) This relation is valid as long as a2≠b2a^2 \neq b^2. If a2=b2a^2=b^2, then (2a2)cos⁡(α−β)=0  ⟹  cos⁡(α−β)=0(2a^2)\cos(\alpha-\beta) = 0 \implies \cos(\alpha-\beta)=0. If cos⁡(α−β)=0\cos(\alpha-\beta)=0, then α−β=(2k+1)π2\alpha-\beta = (2k+1)\frac{\pi}{2}. In this case, cos⁡2α−β2=cos⁡2(kπ2+π4)=12\cos^2\frac{\alpha-\beta}{2} = \cos^2(k\frac{\pi}{2}+\frac{\pi}{4}) = \frac{1}{2}. From (vii): x2a2+y2b2=21+cos⁡(α−β)=21+0=2\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{2}{1+\cos(\alpha-\beta)} = \frac{2}{1+0} = 2. So, x2a2+y2b2=2\frac{x^2}{a^2} + \frac{y^2}{b^2} = 2. If a2=b2a^2=b^2, then x2+y2a2=2  ⟹  x2+y2=2a2\frac{x^2+y^2}{a^2}=2 \implies x^2+y^2=2a^2. Also, a2+b2=a2+a2=2a2a^2+b^2 = a^2+a^2=2a^2. So x2+y2=a2+b2x^2+y^2=a^2+b^2 holds even if a2=b2a^2=b^2.

Let's continue for the general case using (ix). From (v) and (vi): x2=a2cos⁡2α+β2cos⁡2α−β2=a21+cos⁡(α+β)21+cos⁡(α−β)2=a21+cos⁡(α+β)1+cos⁡(α−β)x^2 = a^2 \frac{\cos^2\frac{\alpha+\beta}{2}}{\cos^2\frac{\alpha-\beta}{2}} = a^2 \frac{\frac{1+\cos(\alpha+\beta)}{2}}{\frac{1+\cos(\alpha-\beta)}{2}} = a^2 \frac{1+\cos(\alpha+\beta)}{1+\cos(\alpha-\beta)} y2=b2sin⁡2α+β2cos⁡2α−β2=b21−cos⁡(α+β)21+cos⁡(α−β)2=b21−cos⁡(α+β)1+cos⁡(α−β)y^2 = b^2 \frac{\sin^2\frac{\alpha+\beta}{2}}{\cos^2\frac{\alpha-\beta}{2}} = b^2 \frac{\frac{1-\cos(\alpha+\beta)}{2}}{\frac{1+\cos(\alpha-\beta)}{2}} = b^2 \frac{1-\cos(\alpha+\beta)}{1+\cos(\alpha-\beta)}

Adding these two expressions: x2+y2=a2(1+cos⁡(α+β))+b2(1−cos⁡(α+β))1+cos⁡(α−β)x^2+y^2 = \frac{a^2(1+\cos(\alpha+\beta)) + b^2(1-\cos(\alpha+\beta))}{1+\cos(\alpha-\beta)} x2+y2=a2+a2cos⁡(α+β)+b2−b2cos⁡(α+β)1+cos⁡(α−β)x^2+y^2 = \frac{a^2+a^2\cos(\alpha+\beta) + b^2-b^2\cos(\alpha+\beta)}{1+\cos(\alpha-\beta)} x2+y2=(a2+b2)+(a2−b2)cos⁡(α+β)1+cos⁡(α−β)x^2+y^2 = \frac{(a^2+b^2) + (a^2-b^2)\cos(\alpha+\beta)}{1+\cos(\alpha-\beta)} Substitute cos⁡(α+β)\cos(\alpha+\beta) from (ix): x2+y2=(a2+b2)+(a2−b2)(a2+b2a2−b2cos⁡(α−β))1+cos⁡(α−β)x^2+y^2 = \frac{(a^2+b^2) + (a^2-b^2)\left(\frac{a^2+b^2}{a^2-b^2}\cos(\alpha-\beta)\right)}{1+\cos(\alpha-\beta)} x2+y2=(a2+b2)+(a2+b2)cos⁡(α−β)1+cos⁡(α−β)x^2+y^2 = \frac{(a^2+b^2) + (a^2+b^2)\cos(\alpha-\beta)}{1+\cos(\alpha-\beta)} x2+y2=(a2+b2)(1+cos⁡(α−β))1+cos⁡(α−β)x^2+y^2 = \frac{(a^2+b^2)(1+\cos(\alpha-\beta))}{1+\cos(\alpha-\beta)} Since cos⁡α−β2≠0\cos\frac{\alpha-\beta}{2} \neq 0, 1+cos⁡(α−β)=2cos⁡2α−β2≠01+\cos(\alpha-\beta) = 2\cos^2\frac{\alpha-\beta}{2} \neq 0. So, we can cancel the term (1+cos⁡(α−β))(1+\cos(\alpha-\beta)): x2+y2=a2+b2x^2+y^2 = a^2+b^2 This proves statement B.

Part 3: Prove tan⁡αtan⁡β=b2(x2−a2)a2(y2−b2)\tan \alpha \tan \beta = \frac{b^2(x^2-a^2)}{a^2(y^2-b^2)} From Part 2, we have already established tan⁡αtan⁡β=−b2a2\tan \alpha \tan \beta = -\frac{b^2}{a^2} from the third given condition. Now, let's consider the expression b2(x2−a2)a2(y2−b2)\frac{b^2(x^2-a^2)}{a^2(y^2-b^2)}. We need to show that this expression is equal to −b2a2-\frac{b^2}{a^2}. b2(x2−a2)a2(y2−b2)=−b2a2\frac{b^2(x^2-a^2)}{a^2(y^2-b^2)} = -\frac{b^2}{a^2} Assuming a,b≠0a, b \neq 0, we can divide by b2a2\frac{b^2}{a^2}: x2−a2y2−b2=−1\frac{x^2-a^2}{y^2-b^2} = -1 x2−a2=−(y2−b2)x^2-a^2 = -(y^2-b^2) x2−a2=−y2+b2x^2-a^2 = -y^2+b^2 x2+y2=a2+b2x^2+y^2 = a^2+b^2 Since we have already proven x2+y2=a2+b2x^2+y^2 = a^2+b^2, the first statement is also proven.

The proofs rely on the condition α−β≠2nπ\alpha - \beta \neq 2n\pi, which ensures sin⁡α−β2≠0\sin\frac{\alpha-\beta}{2} \neq 0 and cos⁡α−β2≠0\cos\frac{\alpha-\beta}{2} \neq 0 (otherwise x,yx,y would be infinite from the expressions derived for them). If cos⁡α−β2=0\cos\frac{\alpha-\beta}{2} = 0, then α−β=(2k+1)π\alpha-\beta = (2k+1)\pi, which implies cos⁡(α−β)=−1\cos(\alpha-\beta) = -1, and 1+cos⁡(α−β)=01+\cos(\alpha-\beta)=0, leading to division by zero in the expressions for x2x^2 and y2y^2. However, the initial derivation of XX and YY shows that cos⁡α−β2\cos\frac{\alpha-\beta}{2} is in the denominator. If cos⁡α−β2=0\cos\frac{\alpha-\beta}{2}=0, then XX and YY are undefined, implying xx and yy are undefined, which would make the problem ill-posed. Thus, cos⁡α−β2≠0\cos\frac{\alpha-\beta}{2} \neq 0 is implicitly required for xx and yy to be finite.

Final Answer: The proofs are shown above.

Explanation

Solution

  1. Express xa\frac{x}{a} and yb\frac{y}{b} in terms of trigonometric functions of α+β2\frac{\alpha+\beta}{2} and α−β2\frac{\alpha-\beta}{2} by solving the first two linear equations. xa=cos⁡α+β2cos⁡α−β2\frac{x}{a} = \frac{\cos\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}} and yb=sin⁡α+β2cos⁡α−β2\frac{y}{b} = \frac{\sin\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}}.

  2. Square these expressions and add them to find x2+y2x^2+y^2 in terms of cos⁡(α+β)\cos(\alpha+\beta) and cos⁡(α−β)\cos(\alpha-\beta). x2+y2=a2(1+cos⁡(α+β))+b2(1−cos⁡(α+β))1+cos⁡(α−β)x^2+y^2 = \frac{a^2(1+\cos(\alpha+\beta)) + b^2(1-\cos(\alpha+\beta))}{1+\cos(\alpha-\beta)}.

  3. From the third given condition, derive a relationship between cos⁡(α+β)\cos(\alpha+\beta) and cos⁡(α−β)\cos(\alpha-\beta): tan⁡αtan⁡β=−b2a2  ⟹  (a2+b2)cos⁡(α−β)=(a2−b2)cos⁡(α+β)\tan \alpha \tan \beta = -\frac{b^2}{a^2} \implies (a^2+b^2)\cos(\alpha-\beta) = (a^2-b^2)\cos(\alpha+\beta).

  4. Substitute this relationship into the expression for x2+y2x^2+y^2 to simplify and prove x2+y2=a2+b2x^2+y^2 = a^2+b^2.

  5. To prove the first statement, substitute the proven x2+y2=a2+b2x^2+y^2 = a^2+b^2 into the expression b2(x2−a2)a2(y2−b2)\frac{b^2(x^2-a^2)}{a^2(y^2-b^2)} and show it equals −b2a2-\frac{b^2}{a^2}, which is already known to be tan⁡αtan⁡β\tan \alpha \tan \beta.