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

Last updated date: 21st Mar 2023

•

Total views: 307.5k

•

Views today: 6.86k

Answer

Verified

307.5k+ views

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 $\tan (\alpha + \beta )$.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE