Question

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

Answer 1

Hint: According to given in the question first of all we will take $b\sec \theta $ to the right hand side of the given expression $a\tan \theta + b\sec \theta = c$. After it we will make the both sides square of the obtained equation.

Formula used:
${(a - b)^2} = {a^2} + {b^2} - 2ab.....................(1)$
After solving the obtained equation with the help of the formula (1) we will find the roots for the equation as given in the question that $\alpha $and $\beta $ are distinct roots of the equation $a\tan \theta + b\sec \theta = c$
Now, we can find the roots with the help of the quadratic equation for example as know that,
For a general quadratic equation: $a{x^2} + bx + c = 0$ if m and n are the roots of the quadratic equation then,
Sum of the roots $m + n = - \dfrac{b}{a}$
And the product of the roots $mn = \dfrac{c}{a}$
As we have obtained the value of the roots now we can substitute the obtained root in left hand side of the equation $\tan (\alpha + \beta ) = \dfrac{{2ac}}{{{a^2} - {c^2}}}$ which is $\tan (\alpha + \beta )$ but, before that we will expand the $\tan (\alpha + \beta)$ with the help of the formula given below:
$\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 + \tan a\tan b}}$…………………………………………….(2)
Hence, after substituting the value of the roots in the expansion of $\tan (\alpha + \beta )$ we show that:
$\tan (\alpha + \beta ) = \dfrac{{2ac}}{{{a^2} - {c^2}}}$

Complete step by step answer:
Given,
$\alpha $ and $\beta $ are distinct roots of $a\tan \theta + b\sec \theta = c$
Step 1: First of all we will take $b\sec \theta $ to the right hand side of the given expression $a\tan \theta + b\sec \theta = c$
Hence,
$a\tan \theta - c = - b\sec \theta $
On multiplying with (-) both sides of the equation obtained just above,
$c - a\tan \theta = b\sec \theta $………………..(3)
Step 2: Now, on squaring the both sides of the equation obtained in step (1)
$\Rightarrow {(c - a\tan \theta )^2} = {(b\sec \theta )^2}$
Step 3: Now, to find the square of the obtained equation we will use the formula (1) as mentioned in the solution hint.
$\Rightarrow {c^2} + {a^2}{\tan ^2}\theta - 2ac\tan \theta = {b^2}{\sec ^2}\theta $…………………………(4)
Step 4: As we know that, ${\sec ^2}\theta = 1 + {\tan ^2}\theta $ hence, on substituting the value of ${\sec ^2}\theta $ in obtained equation (4)
$\Rightarrow {c^2} + {a^2}{\tan ^2}\theta - 2ac\tan \theta = {b^2}(1 + {\tan ^2}\theta )$……………………….(5)
Step 5: Now, we will arrange all the terms of $\tan \theta $ form the equation (5)
Hence,
$
\Rightarrow {c^2} + {a^2}{\tan ^2}\theta - 2ac\tan \theta - {b^2}(1 + {\tan ^2}\theta ) = 0 \\
\Rightarrow {\tan ^2}\theta ({a^2} - {b^2}) - 2ac\tan \theta + ({c^2} - {b^2}) = 0..............................(6) \\
 $
Step 6: As given in the question that $\alpha $ and $\beta $ are distinct roots of $a\tan \theta + b\sec \theta = c$ hence, as obtained above we can find the roots from equation (6) as we know that,
$\tan \alpha + \tan \beta = - \dfrac{b}{a}$
So, the sum of the roots: $\tan \alpha + \tan \beta = \dfrac{{2ac}}{{{a^2} - {b^2}}}.........................(7)$
And, $\tan \alpha \tan \beta = \dfrac{c}{a}$
So, the product of the roots:
$\Rightarrow \tan \alpha \tan \beta = \dfrac{{({c^2} - {b^2})}}{{({a^2} - {b^2})}}$…………………………………(8)
Step 7: Now, we have to expand the left-hand side of the given expression which is $\tan (\alpha + \beta )$with the help of the formula (2) as mentioned in the solution hint.
Hence,
$\Rightarrow \tan (\alpha + \beta ) = \dfrac{{\tan \alpha + \tan \beta }}{{1 + \tan \alpha \tan \beta }}$
On substituting the value of sum and products of the roots from (7) and (8),
$\Rightarrow \tan (\alpha + \beta ) = \dfrac{{\dfrac{{2ac}}{{{a^2} - {b^2}}}}}{{1 - \dfrac{{{c^2} - {b^2}}}{{{a^2} - {b^2}}}}}$
On solving the equation obtained,
$
\Rightarrow \tan (\alpha + \beta ) = \dfrac{{\dfrac{{2ac}}{{{a^2} - {b^2}}}}}{{\dfrac{{{a^2} - {b^2} - {c^2} + {b^2}}}{{{a^2} - {b^2}}}}} \\
\Rightarrow \tan (\alpha + \beta ) = \dfrac{{\dfrac{{2ac}}{{{a^2} - {b^2}}}}}{{\dfrac{{{a^2} - {c^2}}}{{{a^2} - {b^2}}}}} \\
\Rightarrow \tan (\alpha + \beta ) = \dfrac{{2ac}}{{{a^2} - {c^2}}} \\
 $
L.H.S. = R.H.S.

Hence, we have proved that $\tan (\alpha + \beta ) = \dfrac{{2ac}}{{{a^2} - {c^2}}}$with the help of the roots $\alpha $and $\beta $ and, the equations obtained.

Note:
As given, $\alpha $and $\beta $ are distinct roots of $a\tan \theta + b\sec \theta = c$ hence, on solving the equation we can find the sum and product of the roots which are $\alpha $ and $\beta $.
To find the roots $\alpha $ and $\beta $ we have to make the given equation $a\tan \theta + b\sec \theta = c$ we have to make it in the form of a quadratic equation because in the question there are two roots $\alpha $ and $\beta $.
After finding the sum and product of roots it is necessary to take care of the signs as if $\alpha $ and $\beta $ are two roots if a quadratic equation then,
$\tan \alpha + \tan \beta = - \dfrac{b}{a}$