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If \(\alpha, \beta\) and \(\gamma\) are angles that satisfy the following conditions, find the value of \(x y z\).
1. \(\tan (\alpha)+\tan (\beta)+\tan (\gamma)\)
\(=\tan (\alpha) \cdot \tan (\beta) \cdot \tan (\gamma)\)
2. \(x=\cos (\alpha)+i \sin (\alpha)\)
3. \(y=\cos (\boldsymbol{\beta})+i \sin (\boldsymbol{\beta})\)
4. \(z=\cos (\gamma)+i \sin (\gamma)\)
Options:
1. \(\tan (\alpha)+\tan (\beta)+\tan (\gamma)\)
\(=\tan (\alpha) \cdot \tan (\beta) \cdot \tan (\gamma)\)
2. \(x=\cos (\alpha)+i \sin (\alpha)\)
3. \(y=\cos (\boldsymbol{\beta})+i \sin (\boldsymbol{\beta})\)
4. \(z=\cos (\gamma)+i \sin (\gamma)\)
Solution:
1501 Upvotes
Verified Answer
The correct answer is:
\pm 1
It is given that,
\(\begin{aligned}
& \tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma \\
& \Rightarrow \tan \alpha+\tan \beta=-\tan \gamma(1-\tan \alpha \tan \beta) \\
& \Rightarrow \quad \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\tan (-\gamma) \\
& \Rightarrow \tan (\alpha+\beta)=\tan (-\gamma) \Rightarrow \alpha+\beta=n \pi-\gamma, n \in \mathbf{I} \\
& \Rightarrow \quad \alpha+\beta+\gamma=n \pi, n \in \mathbf{I} \\
& \text {and } x=\cos \alpha+i \sin \alpha, y=\cos \beta+i \sin \beta, \\
& z=\cos \gamma+i \sin \gamma \\
& \therefore \quad x y z=e^{i \alpha} . e^{i \beta} . e^{i \gamma} \\
& \quad=e^{i(\alpha+\beta+\gamma)}=e^{i n \pi}=\cos n \pi=1 \text { or }-1 \\
& \therefore \quad x y z= \pm 1
\end{aligned}\)
Hence, option (c) is correct.
\(\begin{aligned}
& \tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma \\
& \Rightarrow \tan \alpha+\tan \beta=-\tan \gamma(1-\tan \alpha \tan \beta) \\
& \Rightarrow \quad \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\tan (-\gamma) \\
& \Rightarrow \tan (\alpha+\beta)=\tan (-\gamma) \Rightarrow \alpha+\beta=n \pi-\gamma, n \in \mathbf{I} \\
& \Rightarrow \quad \alpha+\beta+\gamma=n \pi, n \in \mathbf{I} \\
& \text {and } x=\cos \alpha+i \sin \alpha, y=\cos \beta+i \sin \beta, \\
& z=\cos \gamma+i \sin \gamma \\
& \therefore \quad x y z=e^{i \alpha} . e^{i \beta} . e^{i \gamma} \\
& \quad=e^{i(\alpha+\beta+\gamma)}=e^{i n \pi}=\cos n \pi=1 \text { or }-1 \\
& \therefore \quad x y z= \pm 1
\end{aligned}\)
Hence, option (c) is correct.
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