Question

If alpha and beta are the solutions of the equation a tan theta + b sec theta = c then show that
tan ( alpha + beta ) = 2ac/(a^2 - b^2)

please answer this question only and please don't link to any other question. Thank you

Answer 1

↑ Here is your answer ↓
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[tex]atanA + bsecA =c [/tex]

$bsecA = c-atanA$

squaring both sides,

[tex] b^2sec^2A=(c-atanA)^2[/tex]

$b^2(1+tan^2)=c^2+a^2tan^2A - 2actanA$

$(b^2-a^2)tan^2A+2actanA+b^2-c^2=0$

since alpha and beta are the roots of the eq. so,

$tan(alpha)+tan(beta)=-2ac/(b^2-a^2)$

$tan(alpha).tan(beta)=(b^2-c^2)/(b^2-a^2)$

$tan(alpha+beta)={tan(alpha)+tan(beta)}/{1-tan(alpha).tan(beta)}$

$=[-2ac/(b^2-a^2)]/[1-(b^2-c^2)/(b^2-a^2)]$

$=-2ac/(c^2-a^2)$

$=2ac/(a^2-c^2)$

Answer 2

it is helpful for you..... plz read the pic..