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Prove that

$${ \left( \frac { \cos (\alpha +\beta ) }{ \cos (\alpha -\beta ) } -\frac { \cos (\alpha +\gamma ) }{ \cos (\alpha -\gamma ) } \right) }^{ 2 }+{ \left( \frac { \sin (\alpha +\beta ) }{ \cos (\alpha -\beta ) } -\frac { \sin (\alpha +\gamma ) }{ \cos (\alpha -\gamma ) } \right) }^{ 2 }={ \left( \sec ^{ 2 }{ (\alpha -\beta ) } \sin ^{ 2 }{ (\beta -\gamma ) } \sec ^{ 2 }{ (\gamma -\alpha ) } \right) }$$

I brought everything to the common denominator noticing bot the fractions had the same denominator but after that I find I am stick with a mess

sidt36
  • 261

3 Answers3

1

$$\dfrac{\cos^2(\alpha+\beta)+\sin^2(\alpha+\beta)}{\cos^2(\alpha-\beta)}+\dfrac{\cos^2(\alpha+\gamma)+\sin^2(\alpha+\gamma)}{\cos^2(\alpha-\gamma)}$$ $$-2\cdot\dfrac{\cos(\alpha+\beta)\cos(\alpha+\gamma)+\sin(\alpha+\beta)\sin(\alpha+\gamma)}{\cos(\alpha-\beta)\cos(\alpha-\gamma)}$$

$$=\dfrac1{\cos^2(\alpha-\beta)}+\dfrac1{\cos^2(\alpha-\gamma)}-\dfrac{2\cos\{\alpha+\beta-(\alpha+\gamma)\}}{\cos(\alpha-\beta)\cos(\alpha-\gamma)}$$

$$=\dfrac{\cos^2(\alpha-\gamma)+\cos^2(\alpha-\beta)-2\cos(\alpha-\beta)\cos(\alpha-\gamma)\cos(\beta-\gamma)}{\cos^2(\alpha-\beta)\cos^2(\alpha-\gamma)}$$

Now using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ for the numerator $$\cos^2(\alpha-\gamma)+\cos^2(\alpha-\beta)-2\cos(\alpha-\beta)\cos(\alpha-\gamma)\cos(\beta-\gamma)$$

$$=1+\cos^2(\alpha-\gamma)-\sin^2(\alpha-\beta)-2\cos(\alpha-\beta)\cos(\alpha-\gamma)\cos(\beta-\gamma)$$

$$=1+\cos(2\alpha-\beta-\gamma)\cos(\beta-\gamma)-2\cos(\alpha-\beta)\cos(\alpha-\gamma)\cos(\beta-\gamma)$$

$$=1+\cos(\beta-\gamma)\{\cos(2\alpha-\beta-\gamma)-2\cos(\alpha-\beta)\cos(\alpha-\gamma)\}$$

Now use Werner Formulas $2\cos A\cos B=\cdots$ on $$2\cos(\alpha-\beta)\cos(\alpha-\gamma)$$

Community
  • 1
lab bhattacharjee
  • 274,582
1

\begin{align*} \frac { \cos (\alpha +\beta ) }{ \cos (\alpha -\beta ) } &-\frac { \cos (\alpha +\gamma ) }{ \cos (\alpha -\gamma ) } = \frac{\cos (\alpha +\beta )\cos (\alpha -\gamma ) - \cos (\alpha +\gamma )\cos (\alpha -\beta )}{\cos (\alpha -\beta )\cos (\alpha -\gamma )} \\ &= \frac{1}{2}\left(\frac{( \cos(2\alpha + \beta - \gamma) + \cos(\beta + \gamma) - (\cos(2\alpha - \beta + \gamma) + \cos(\beta + \gamma) )}{\cos (\alpha -\beta )\cos (\alpha -\gamma )}\right)\\ &= \frac{1}{2}\left(\frac{( \cos(2\alpha + \beta - \gamma) - \cos(2\alpha - \beta + \gamma) }{\cos (\alpha -\beta )\cos (\alpha -\gamma )}\right)\\ &= \frac{1}{2}\frac{2\sin(2\alpha)\sin(\gamma-\beta)}{\cos (\alpha -\beta )\cos (\alpha -\gamma )}\\ &=\frac{\sin(2\alpha)\sin(\gamma-\beta)}{\cos (\alpha -\beta )\cos (\alpha -\gamma )} \end{align*} Similarly, \begin{align*} \left( \frac { \sin (\alpha +\beta ) }{ \cos (\alpha -\beta ) } -\frac { \sin (\alpha +\gamma ) }{ \cos (\alpha -\gamma ) } \right) &=\frac{\cos(2\alpha)\sin(\beta-\gamma)}{\cos (\alpha -\beta )\cos (\alpha -\gamma )} \end{align*} Squaring and adding, we get \begin{align*} \frac{\sin^2(\beta-\gamma)}{\cos^2(\alpha -\beta )\cos^2(\alpha -\gamma )} &= \sec^2(\alpha -\beta )\sec^2(\alpha -\gamma )\sin^2(\beta-\gamma) \end{align*} Note that the given equality as stated in the problem is false when $\alpha = \beta = \gamma$.

0

We easily get LHS $ = \sec^2 (\alpha - \beta)+\sec^2 (\alpha-\gamma) - \dfrac{2\cos (\beta -\gamma)}{\cos (\alpha - \beta) \cos (\alpha - \gamma)}$

$= \sec^2 (\alpha - \beta)+\sec^2 (\alpha-\gamma) - \dfrac{2\cos (\beta -\alpha+\alpha-\gamma)}{\cos (\alpha - \beta) \cos (\alpha - \gamma)}$

$=\sec^2 (\alpha - \beta)+\sec^2 (\alpha-\gamma) -2 + 2\tan (\beta-\alpha) \tan (\alpha-\gamma)$

$=\tan^2 (\alpha - \beta)+\tan^2 (\alpha-\gamma) + 2\tan (\beta-\alpha) \tan (\alpha-\gamma)$

$=\left(\tan (\beta-\alpha)+\tan (\alpha-\gamma)\right)^2$

Now using $\tan A + \tan B = \dfrac{\sin (A+B)}{\cos A \cos B}$ we get further

$\left(\dfrac{\sin (\beta - \gamma)}{\cos (\beta -\alpha) \cos (\alpha-\gamma)}\right)^2$

$=\sec^2 (\beta -\alpha) \sin^2 (\beta - \gamma) \sec^2 (\alpha-\gamma)$ = RHS

Hari Shankar
  • 3,636