# If $\alpha $ and $\beta $ are the solutions of the equation $a\tan \theta + b\sec \theta = c$, then show that$\tan (\alpha + \beta ) = \dfrac{{2ac}}{{({a^2} - {c^{^2}})}}$.

Answer

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Hint: Here, we will use the trigonometric formulae and roots of quadratic equation formulas.

Given,

$a\tan \theta + b\sec \theta = c \to (1)$

Now, equation (1) can be written as

$ \Rightarrow b\sec \theta = c - a\tan \theta $

Squaring the equation on both sides, we get

$\begin{gathered}

\Rightarrow {b^2}{\sec ^2}\theta = {(c - a\tan \theta )^2} \\

\Rightarrow {b^2} (1 + {\tan ^2}\theta ) = {c^2} + {a^2}{\tan ^2}\theta - 2ac\tan \theta [\because {\sec ^2}\theta = 1 + {\tan ^2}\theta ] \\

\Rightarrow ({a^2} - {b^2}){\tan ^2}\theta - 2ac\tan \theta + ({c^2} - {b^2}) = 0 \to (2) \\

\end{gathered} $

As, we can see equation (2) is in the form of quadratic equation of $\tan \theta $.Now, let $\tan \alpha $ and $\tan \beta $ be the roots of equation (2) (since, $\alpha $ and $\beta $ are roots of the equation).Therefore, we can apply the sum of roots and product of roots formula such that we get

$\tan \alpha + \tan \beta = \dfrac{{2ac}}{{({a^2} - {b^2})}} \to (3)[\because $Sum of the roots =$\dfrac{{{\text{ - coefficient of x}}}}{{{\text{coefficient of}}{{\text{x}}^{\text{2}}}}}$]

$\tan \alpha \tan \beta = \dfrac{{({c^2} - {b^2})}}{{({a^2} - {b^2})}} \to (4)[\because $Product of the roots =$\dfrac{{{\text{coefficient constant}}}}{{{\text{coefficient of}}{{\text{x}}^{\text{2}}}}}$]

Now, we need to find \[\tan (\alpha + \beta )\], we know that

\[\tan (\alpha + \beta ) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha .\tan \beta }} \to (5)\].

So, substituting equation (3) and (4) in equation (5), we get

\[\begin{gathered}

\Rightarrow \tan (\alpha + \beta ) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha .\tan \beta }} \\

\Rightarrow \tan (\alpha + \beta ) = \dfrac{{\dfrac{{2ac}}{{({a^2} - {b^2})}}}}{{1 - (\dfrac{{({c^2} - {b^2})}}{{({a^2} - {b^2})}})}} \\

\Rightarrow \tan (\alpha + \beta ) = \dfrac{{2ac}}{{({a^2} - {b^2}) - ({c^2} - {b^2})}} \\

\Rightarrow \tan (\alpha + \beta ) = \dfrac{{2ac}}{{({a^2} - {c^{^2}})}} \\

\end{gathered} \]

Hence, we proved that $\tan (\alpha + \beta ) = \dfrac{{2ac}}{{({a^2} - {c^{^2}})}}$.

Note: Here, we need to convert the given equation in form of $\tan \theta $ such that we can use $\tan \alpha $ and $\tan \beta $ as the roots of the equation to find $\tan (\alpha + \beta )$.

Given,

$a\tan \theta + b\sec \theta = c \to (1)$

Now, equation (1) can be written as

$ \Rightarrow b\sec \theta = c - a\tan \theta $

Squaring the equation on both sides, we get

$\begin{gathered}

\Rightarrow {b^2}{\sec ^2}\theta = {(c - a\tan \theta )^2} \\

\Rightarrow {b^2} (1 + {\tan ^2}\theta ) = {c^2} + {a^2}{\tan ^2}\theta - 2ac\tan \theta [\because {\sec ^2}\theta = 1 + {\tan ^2}\theta ] \\

\Rightarrow ({a^2} - {b^2}){\tan ^2}\theta - 2ac\tan \theta + ({c^2} - {b^2}) = 0 \to (2) \\

\end{gathered} $

As, we can see equation (2) is in the form of quadratic equation of $\tan \theta $.Now, let $\tan \alpha $ and $\tan \beta $ be the roots of equation (2) (since, $\alpha $ and $\beta $ are roots of the equation).Therefore, we can apply the sum of roots and product of roots formula such that we get

$\tan \alpha + \tan \beta = \dfrac{{2ac}}{{({a^2} - {b^2})}} \to (3)[\because $Sum of the roots =$\dfrac{{{\text{ - coefficient of x}}}}{{{\text{coefficient of}}{{\text{x}}^{\text{2}}}}}$]

$\tan \alpha \tan \beta = \dfrac{{({c^2} - {b^2})}}{{({a^2} - {b^2})}} \to (4)[\because $Product of the roots =$\dfrac{{{\text{coefficient constant}}}}{{{\text{coefficient of}}{{\text{x}}^{\text{2}}}}}$]

Now, we need to find \[\tan (\alpha + \beta )\], we know that

\[\tan (\alpha + \beta ) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha .\tan \beta }} \to (5)\].

So, substituting equation (3) and (4) in equation (5), we get

\[\begin{gathered}

\Rightarrow \tan (\alpha + \beta ) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha .\tan \beta }} \\

\Rightarrow \tan (\alpha + \beta ) = \dfrac{{\dfrac{{2ac}}{{({a^2} - {b^2})}}}}{{1 - (\dfrac{{({c^2} - {b^2})}}{{({a^2} - {b^2})}})}} \\

\Rightarrow \tan (\alpha + \beta ) = \dfrac{{2ac}}{{({a^2} - {b^2}) - ({c^2} - {b^2})}} \\

\Rightarrow \tan (\alpha + \beta ) = \dfrac{{2ac}}{{({a^2} - {c^{^2}})}} \\

\end{gathered} \]

Hence, we proved that $\tan (\alpha + \beta ) = \dfrac{{2ac}}{{({a^2} - {c^{^2}})}}$.

Note: Here, we need to convert the given equation in form of $\tan \theta $ such that we can use $\tan \alpha $ and $\tan \beta $ as the roots of the equation to find $\tan (\alpha + \beta )$.

Last updated date: 03rd Oct 2023

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