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Prove the following.
If cos⁻¹ x/a + cos⁻¹ y/b = α, prove that (x²)/(a²) - (2xy)/(ab) cos α + (y²)/(b²) = sin² α
If $\cos^{-1} \frac{x}{a} + \cos^{-1} \frac{y}{b} = \alpha$, prove that $\frac{x^2}{a^2} - \frac{2xy}{ab} \cos \alpha + \frac{y^2}{b^2} = \sin^2 \alpha$
Given: $\cos^{-1} \frac{x}{a} + \cos^{-1} \frac{y}{b} = \alpha$
To show: $\frac{x^2}{a^2} - \frac{2xy}{ab} \cos \alpha + \frac{y^2}{b^2} = \sin^2 \alpha$
Answer
Answer
Let $\cos^{-1}\frac{x}{a} = \theta$, so $\cos\theta = \frac{x}{a}$ and $\sin\theta = \sqrt{1 - \frac{x^2}{a^2}}$.
- Let $\cos^{-1}\frac{x}{a} = \theta$; then $\cos\theta = \frac{x}{a}$ and $\sin\theta = \sqrt{1 - \frac{x^2}{a^2}}$ (using $\sin^2\theta + \cos^2\theta = 1$).
- The given relation gives $\cos^{-1}\frac{y}{b} = \alpha - \theta$, so $\frac{y}{b} = \cos(\alpha - \theta)$.
- Expanding by $\cos(A - B) = \cos A\cos B + \sin A\sin B$: $\frac{y}{b} = \cos\alpha \cdot \frac{x}{a} + \sin\alpha\sqrt{1 - \frac{x^2}{a^2}}$.
- Rearranging: $\frac{y}{b} - \frac{x}{a}\cos\alpha = \sin\alpha\sqrt{1 - \frac{x^2}{a^2}}$.
- Squaring both sides: $\frac{y^2}{b^2} - \frac{2xy}{ab}\cos\alpha + \frac{x^2}{a^2}\cos^2\alpha = \sin^2\alpha\left(1 - \frac{x^2}{a^2}\right)$.
- Bringing the $x^2$ terms together: $\frac{x^2}{a^2}(\cos^2\alpha + \sin^2\alpha) - \frac{2xy}{ab}\cos\alpha + \frac{y^2}{b^2} = \sin^2\alpha$.
- Since $\cos^2\alpha + \sin^2\alpha = 1$: $\frac{x^2}{a^2} - \frac{2xy}{ab}\cos\alpha + \frac{y^2}{b^2} = \sin^2\alpha$. Hence proved.