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If $\alpha $ and $\beta$ are roots of equation $a\tan\theta +b \sec\theta=c$. Prove that $\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$

i have converted tan to sin and cos and reached to $\sin^2\theta(a^2-c^2+2ac) + c^2-b^2-ac=0$. how do i proceed

thanks

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    See here: http://math.stackexchange.com/questions/838082 – 2017-01-28
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    i want to know do i proceed in my solution.thanks btw – 2017-01-28
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    I think there is a mistake in your computations. – 2017-01-28

3 Answers 3

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I think there is mistake in your computations. $$(a\sin\theta+b)^2=c^2(1-\sin^2\theta),$$ which does not give, what you wish.

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We have $a\tan\theta +b \sec\theta=c$. So,

\begin{align*} c-a\tan \theta & = b \sec \theta\\ (c-a\tan \theta)^2 & = (b\sec \theta)^2\\ (a^2-b^2)\tan^2 \theta -2ac \tan \theta+(c^2-b^2) & = 0. \end{align*} This is a quadratic in $\tan \theta$. It has two solutions $\tan \alpha$ and $\tan \beta$. So \begin{align*} \tan \alpha + \tan \beta & = \frac{2ac}{a^2-b^2}\\ \tan \alpha \cdot \tan \beta & = \frac{c^2-b^2}{a^2-b^2} \end{align*} Now, $$\tan(\alpha+\beta)=\frac{\tan \alpha +\tan\beta}{1-\tan \alpha \tan \beta}$$ $$\tan(\alpha+\beta)=\frac{\frac{2ac}{a^2-b^2}}{1-\frac{c^2-b^2}{a^2-b^2}}$$ $$\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$$

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We can write the quadratic equation as: $$a\tan \theta + b\sec \theta = c $$ $$\Rightarrow a\tan \theta + b\sqrt {\tan^2 \theta + 1} = c $$ $$\Rightarrow (a^2-b^2)\tan^2 \theta -2ac \tan \theta +(c^2-b^2)=0$$

Now $$\tan (\alpha +\beta ) =\frac {\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}. $$ Can you take it from here? Hope it helps.

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    We have a direct formula for $\tan (\alpha + \beta) $. So it is easy to keep the equation of yours in $\tan $ form itself. – 2017-01-28