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If \alpha, \beta, \gamma do not differ by a multiple of \pi and if \frac{\cos (\alpha+\theta)}{\sin (\beta+\gamma)}=\frac{\cos (\beta+\theta)}{\sin (\gamma+x)}=\frac{\cos (\gamma+\theta)}{\sin (\alpha+\beta)}=k

Then k equals 

 

Option: 1

\pm 2


Option: 2

\pm 1 / 2


Option: 3

0


Option: 4

\pm1


Answers (1)

best_answer

Taking first two members, then 

\begin{aligned} & 2 \cos (\alpha+\theta) \sin (\gamma+\alpha) \\ \\&= 2 \cos (\beta+\theta) \sin (\beta+\gamma) \\ \\& \Rightarrow \quad \sin (2 \alpha+\theta+\gamma)-\sin (\theta-\gamma) \\ \\&= \sin (2 \beta+\theta+\gamma)-\sin (\theta-\gamma) \end{aligned}

\begin{aligned} & \Rightarrow \quad \sin (2 \alpha+\theta+\gamma)=\sin (2 \beta+\theta+\gamma) \\ \\& \Rightarrow \quad 2 \alpha+\theta+\gamma=n \pi+(-1)^n(2 \beta+\theta+\gamma) \\ \\& \text { for } n=0,2 \alpha+\theta+\gamma=2 \beta+\theta+\gamma \end{aligned}

\therefore \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \alpha =\beta

Similarly taking last two members, we get

                        \beta =\gamma

Hence,            \alpha =\beta =\gamma

Also ,take         n=1,-1

Then , we get  \alpha+\beta+\gamma+\theta=\frac{\pi}{2},-\frac{\pi}{2}

\begin{aligned} \therefore \quad k & =\frac{\cos (\alpha+\theta)}{\sin (\beta+\gamma)}=\frac{\cos \left(\frac{\pi}{2}-(\beta+\gamma)\right)}{\sin (\beta+\gamma)} \\ \\& =\frac{\sin (\beta+\gamma)}{\sin (\beta+\gamma)}=1 \end{aligned}

\\k=\frac{\cos \left(-\frac{\pi}{2}-(\beta+\gamma)\right)}{\sin (\beta+\gamma)}=-1 \\\\Hence, \quad k= \pm 1

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Gaurav

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