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Question: If \(\alpha \) and \(\beta \) are distinct roots of \(a\tan \theta + b\sec \theta = c,\) then show t...

If α\alpha and β\beta are distinct roots of atan⁡θ+bsec⁡θ=c,a\tan \theta + b\sec \theta = c, then show that, tan⁡(α+β)=2aca2−c2\tan (\alpha + \beta ) = \dfrac{{2ac}}{{{a^2} - {c^2}}}

Explanation

Solution

According to given in the question first of all we will take bsec⁡θb\sec \theta to the right hand side of the given expression atan⁡θ+bsec⁡θ=ca\tan \theta + b\sec \theta = c. After it we will make the both sides square of the obtained equation.

Formula used:
(a−b)2=a2+b2−2ab.....................(1){(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 α\alpha and β\beta are distinct roots of the equation atan⁡θ+bsec⁡θ=ca\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: ax2+bx+c=0a{x^2} + bx + c = 0 if m and n are the roots of the quadratic equation then,
Sum of the roots m+n=−bam + n = - \dfrac{b}{a}
And the product of the roots mn=camn = \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⁡(α+β)=2aca2−c2\tan (\alpha + \beta ) = \dfrac{{2ac}}{{{a^2} - {c^2}}} which is tan⁡(α+β)\tan (\alpha + \beta ) but, before that we will expand the tan⁡(α+β)\tan (\alpha + \beta) with the help of the formula given below:
tan⁡(a+b)=tan⁡a+tan⁡b1+tan⁡atan⁡b\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⁡(α+β)\tan (\alpha + \beta ) we show that:
tan⁡(α+β)=2aca2−c2\tan (\alpha + \beta ) = \dfrac{{2ac}}{{{a^2} - {c^2}}}

Complete step by step answer:
Given,
α\alpha and β\beta are distinct roots of atan⁡θ+bsec⁡θ=ca\tan \theta + b\sec \theta = c
Step 1: First of all we will take bsec⁡θb\sec \theta to the right hand side of the given expression atan⁡θ+bsec⁡θ=ca\tan \theta + b\sec \theta = c
Hence,
atan⁡θ−c=−bsec⁡θa\tan \theta - c = - b\sec \theta
On multiplying with (-) both sides of the equation obtained just above,
c−atan⁡θ=bsec⁡θc - a\tan \theta = b\sec \theta ………………..(3)
Step 2: Now, on squaring the both sides of the equation obtained in step (1)
⇒(c−atan⁡θ)2=(bsec⁡θ)2\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.
⇒c2+a2tan⁡2θ−2actan⁡θ=b2sec⁡2θ\Rightarrow {c^2} + {a^2}{\tan ^2}\theta - 2ac\tan \theta = {b^2}{\sec ^2}\theta …………………………(4)
Step 4: As we know that, sec⁡2θ=1+tan⁡2θ{\sec ^2}\theta = 1 + {\tan ^2}\theta hence, on substituting the value of sec⁡2θ{\sec ^2}\theta in obtained equation (4)
⇒c2+a2tan⁡2θ−2actan⁡θ=b2(1+tan⁡2θ)\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⁡θ\tan \theta form the equation (5)
Hence,
⇒c2+a2tan⁡2θ−2actan⁡θ−b2(1+tan⁡2θ)=0 ⇒tan⁡2θ(a2−b2)−2actan⁡θ+(c2−b2)=0..............................(6)  \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 α\alpha and β\beta are distinct roots of atan⁡θ+bsec⁡θ=ca\tan \theta + b\sec \theta = c hence, as obtained above we can find the roots from equation (6) as we know that,
tan⁡α+tan⁡β=−ba\tan \alpha + \tan \beta = - \dfrac{b}{a}
So, the sum of the roots: tan⁡α+tan⁡β=2aca2−b2.........................(7)\tan \alpha + \tan \beta = \dfrac{{2ac}}{{{a^2} - {b^2}}}.........................(7)
And, tan⁡αtan⁡β=ca\tan \alpha \tan \beta = \dfrac{c}{a}
So, the product of the roots:
⇒tan⁡αtan⁡β=(c2−b2)(a2−b2)\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⁡(α+β)\tan (\alpha + \beta )with the help of the formula (2) as mentioned in the solution hint.
Hence,
⇒tan⁡(α+β)=tan⁡α+tan⁡β1+tan⁡αtan⁡β\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),
⇒tan⁡(α+β)=2aca2−b21−c2−b2a2−b2\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,
⇒tan⁡(α+β)=2aca2−b2a2−b2−c2+b2a2−b2 ⇒tan⁡(α+β)=2aca2−b2a2−c2a2−b2 ⇒tan⁡(α+β)=2aca2−c2  \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⁡(α+β)=2aca2−c2\tan (\alpha + \beta ) = \dfrac{{2ac}}{{{a^2} - {c^2}}}with the help of the roots α\alpha and β\beta and, the equations obtained.

Note:
As given, α\alpha and β\beta are distinct roots of atan⁡θ+bsec⁡θ=ca\tan \theta + b\sec \theta = c hence, on solving the equation we can find the sum and product of the roots which are α\alpha and β\beta .
To find the roots α\alpha and β\beta we have to make the given equation atan⁡θ+bsec⁡θ=ca\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 α\alpha and β\beta .
After finding the sum and product of roots it is necessary to take care of the signs as if α\alpha and β\beta are two roots if a quadratic equation then,
tan⁡α+tan⁡β=−ba\tan \alpha + \tan \beta = - \dfrac{b}{a}