# If $\frac {x}{a}=cos (\theta -\alpha)$.. Hello anyone there. Please help to solve.

If $\frac {x}{a}=cos(\theta -\alpha)$ and $\frac {y}{b}=cot(\theta -\beta)$, prove that: $\frac {x^2}{a^2} -\frac {2xy}{ab} cos(\alpha-\beta)+\frac {y^2}{b^2}=\sin^2(\alpha-\beta)$

My Attempt

$$L.H.S=\frac {x^2}{a^2} - \frac {2xy}{ab} cos(\alpha-\beta)+\frac {y^2}{b^2}$$ $$=cos^2(\theta -\alpha)-2cos(\theta-\alpha).cot(\theta-\beta).cos(\alpha-\beta)+cot^2(\theta-\beta)$$ $$=cos^2(\theta-\alpha)-cos(\theta+\beta-2\alpha).cot(\theta-\beta)-sin(\theta-\beta)+cot^2(\theta-\beta)$$.

• I take it that the "$\frac{x^2}{y^2}$" term is a typo for "$\frac{x^2}{a^2}$" (as in your attempt). There must be another error, however. For instance, if we take $\theta = \pi/2$ and $\alpha = \beta = \pi/4$, then we have $$\frac{x}{a} = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad\frac{y}{b} = \cot\frac{\pi}{4} = 1, \quad \cos(\alpha-\beta) = 1, \quad \sin^2(\alpha-\beta) = 0$$ so that the equation asserts $$\frac{1}{2} - 2\cdot \frac{\sqrt{2}}{2}\cdot 1 \cdot 1 + 1 = 0 \qquad\to\qquad \frac{3}{2} = \sqrt{2}$$ which is obviously(?) untrue. – Blue Oct 9 '16 at 14:19
• How can you assume: $\alpha=\beta$? – pi-π Oct 9 '16 at 15:32
• He showed a counterexample of your claim. – Theorem Oct 9 '16 at 15:54
• If the left-hand side actually simplifies to the right-hand side, then it must do so for all particular values of $\alpha$, $\beta$, $\theta$. Consequently, finding sample values for which the equation fails shows that the simplification can't work in general. If I believed that $(p+q)^2$ always equals $p^2 + q^2$, then you could prove me wrong with a simple counterexample, such as $p=q=1$. – Blue Oct 9 '16 at 15:58
• @Blue, It means the question is wrong. Then, could you please correct it? – pi-π Oct 9 '16 at 16:04

$\cot$ shouls be $\cos$
$$\cos(\alpha-\beta)=\cos\{(\theta-\beta)-(\theta-\alpha)\}=?$$
$$\iff\cos(\alpha-\beta)-\cos(\theta-\beta)\cos(\theta-\alpha)=\sin(\theta-\beta)\sin(\theta-\alpha)$$ Squaring we get $$\{\cos(\alpha-\beta)-\cos(\theta-\beta)\cos(\theta-\alpha)\}^2=\{1-\cos^2(\theta-\beta)\}\{1-\cos^2(\theta-\alpha)\}$$
Replace the values of $\cos(\theta-\beta),\cos(\theta-\alpha)$ & simplify.
• @user354073, Have u noticed $\cos(A-B)$ expansion? Then replacement of $\sin^2$ with $1-\cos^2$ – lab bhattacharjee Oct 10 '16 at 16:39