Math, asked by hazarikabishnu429, 2 months ago

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


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Answers

Answered by mathdude500
5

 \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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