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If \alpha and \beta are the solutions of the equation 2^{\log z_2(\cos 2 \theta+b \operatorname{cosec} 2 \theta)}=c, then \tan (3 \alpha+3 \beta) is equal to

Option: 1

\frac{a}{c}\left(\frac{a^2-3 c^2}{c^2-3 a^2}\right)


Option: 2

\frac{a}{c}\left(\frac{c^2-3 a^2}{a^2-3 c^2}\right)


Option: 3

\frac{c}{a}\left(\frac{c^2-3 a^2}{a^2-3 c^2}\right)


Option: 4

None of these


Answers (1)

best_answer

 Given equation is 2^{\log _2(a \cot 2 \theta+b \operatorname{cosec} 2 \theta)}=c

\begin{aligned} & \therefore \quad a \cot 2 \theta+b \operatorname{cosec} 2 \theta=c \\ & \Rightarrow \quad a\left(\frac{1-\tan ^2 \theta}{2 \tan \theta}\right)+b\left(\frac{1+\tan ^2 \theta}{2 \tan \theta}\right)=c \\ & \therefore \quad(b-a) \tan ^2 \theta-2 c \tan \theta+(a+b)=0 \\ & \end{aligned}

If \alpha and \beta are the roots of the above equation, then
                \tan \alpha+\tan \beta=\frac{2 c}{b-a}
and          \tan \beta =\frac{a+b}{b-a}
\therefore \quad \quad \tan (\alpha+\beta) =\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{-c}{a}
Hence ,       \tan (3 \alpha+3 \beta) =\tan 3(\alpha+\beta)
                  \begin{aligned} \\ & =\frac{3 \tan (\alpha+\beta)-\tan ^3(\alpha+\beta)}{1-3 \tan ^2(\alpha+\beta)} \\ & =\frac{3\left(\frac{-c}{a}\right)-\left(\frac{-c}{a}\right)^3}{1-3\left(\frac{-c}{a}\right)^2}=\frac{c\left(c^2-3 a^2\right)}{a\left(a^2-3 c^2\right)} \end{aligned}

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