Question

if alpha and beta are the roots of the equation atanA + bsecA=c then prove that tan(alpha +beta )=-2ac/ b²-a²


i need the complete explanation​

Answer 1

$\large\underline\blue{\bold{Given \: Question \: (correct \: statement) :- }}$

• if alpha and beta are the roots of the equation atanA + bsecA = c, then prove that tan(alpha +beta )=-2ac/ a²-c²

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$\bf \:\large \red{AηsωeR :.} ✍$

$\large\underline\blue{\bold{Given :- }}$

• alpha and beta are the roots of the equation : atanA + bsecA = c

$\large \red{\bf \:To  \: prove :- }$

• tan(alpha +beta )=-2ac/ a²- c²

$\large\underline\purple{\bold{Solution :- }}$

▪︎Let alpha = x and beta = y

● So, x and y are the roots of the equation :

a tanA + b secA = c.

☆Now,

$\bf \:atanA + bsecA = c$

$\bf \:  ⟼ bsecA = c - atanA$

$\bf \:\large \blue{squaring \: both \: sides \: we \: get}$

$\bf \:  ⟼ {(bsecA)}^{2} = {(c - atanA)}^{2}$

$\bf \:  ⟼ {b}^{2} {(secA)}^{2} = {c}^{2} + {a}^{2} {(tanA)}^{2} - 2actanA$

$\bf \:  ⟼ {b}^{2} (1 + {tan}^{2} A) = {c}^{2} + {a}^{2} {tan}^{2} A - 2actanA$

$\bf \:  ⟼ {b}^{2} + {b}^{2} {tan}^{2} A = {c}^{2} + {a}^{2} {tan}^{2} A - 2actanA$

$\bf \:  ⟼ ( {a}^{2} - {b}^{2} ) {tan}^{2} A - 2ac \: tanA + {c}^{2} - {b}^{2} = 0$

☆Now, its a quadratic equation in tanA,

☆So, it implies equation having two roots tanx and tany.

☆So, it further implies,

☆Sum of roots,

$\bf \:  ⟼ tanx + tany = - \dfrac{ - 2ac}{ {a}^{2} - {b}^{2} }$

$\bf\implies \:tanx + tany = \dfrac{2ac}{ {a}^{2} - {b}^{2} }$

☆and product of roots,

$\bf \:  ⟼ tanx \: tany = \dfrac{ {c}^{2} - {b}^{2} }{ {a}^{2} - {b}^{2} }$

☆ Now, we know

$\bf \:  ⟼ tan(x + y) = \dfrac{tanx + tany}{1 - tanx \: tany}$

$\bf \:  ⟼ tan(x + y) = \dfrac{\dfrac{2ac}{ {a}^{2} - {b}^{2} } }{1 - \dfrac{ {c}^{2} - {b}^{2} }{ {a}^{2} - {b}^{2} } }$

$\bf \:  ⟼ tan(x + y) = \dfrac{2ac}{ {a}^{2} - {b}^{2} - {c}^{2} + {b}^{2} }$

$\bf \:  ⟼ tan(x + y) = \dfrac{2ac}{ {a}^{2} - {c}^{2} }$

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